Latent Growth Curve modeling - some Q... PreviousNext
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 Lee_50 posted on Friday, December 30, 2005 - 12:46 pm
I use the Mplus User's Guide and then have some questions.

1. Some papers write down the status facotr, it is the same the intercept growth factor, isn't it? ( I think it is same but I want to make sure.)

2. Fix the model, consist of two basic random-effects growth factors: an overall status factor and a linear growth facotr...so overall status factor is also the same in (1).

3. Want to use: Full-information maximum-likelihood estimation under missingness, so I search it but the option='estimator', it is not find...could you tell me? just default or ML( is the FIML)

4. Want to use: satorra-bentler corrections to chi-square tests of model fit and parameter standard errors, but if I just use the Multilevel model, I got chi-square test value. is this same??

5. Use the other kind of modelin, I got the P-value from output, but the mulitlevel model result didn't give this part(p-value) how to calcuate p-value? or it is possible in Mplus?

Some questions is not difficult. But I really want to make sure. Cuz, each book and paper..they use the different word even if the same meanning..this kinds of thing it is not good me..confuse.
Thank you so much for helping. It is too help to me.
 Linda K. Muthen posted on Saturday, December 31, 2005 - 9:37 am
1-2. When the slope growth factor has the zero time score at the first timepoint, the intercept growth factor describes initial status. If the zero time score is at the last timepoint, the intercept growth factor describes final status. Perhaps this is what you mean.

3. Any maximum likelihood estimator will give you full-information maximum likelihood.

4. MLM gives the Satorra-Bentler chi-square.

5. I think you mean you are not getting chi-square and the p-value for chi-square. When means, variances, and covariances are not sufficient statistics for model estimation, chi-square is not available.
 Geertje Leflot posted on Tuesday, February 24, 2009 - 4:53 am
I have run a latent growth curve model to analyze the development of ADH problems. I have analyzed whether the development of these problems is different in the control and intervention condition, by regressing the intercept and the slope of the growth curve model on the intervention status (cond). There is a significant effect of the intervention status on the slope (not on the intercept).
QUESTION: I would like to visualize the growth of ADH problems in the control and intervention condition. Can this be done in Mplus?

I have already tried this, but I cannot find a way to visualize the growth in the two groups:
Iadh Sadh | ADHD1@0 ADHD2@.5 ADHD3@1 ADHD4@1.5;

Iadh Sadh;
Iadh with Sadh;

Iadh on male cond;
Sadh on male cond;


output: sampstat modindices(3) stand cinterval
TECH4 TECH1;

PLOT: TYPE IS PLOT3;
SERIES is adhd1 - adhd4 (Sadh);
 Linda K. Muthen posted on Tuesday, February 24, 2009 - 9:22 am
Look at the adjusted means plot. I think that will give you what you want.
 Lisa M. Yarnell posted on Thursday, September 29, 2011 - 4:47 pm
Hi Linda, I am running a latent growth model with 4 time points, and received this message: THE MODEL ESTIMATION TERMINATED NORMALLY WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES. CHECK THE TECH4 OUTPUT FOR MORE INFORMATION. PROBLEM INVOLVING VARIABLE SLOPE.

Do you see anything wrong with my code below? I tried loosening starting values for loadings to slope, and tried providing a positive starting value for the variance of SLOPE, since the message suggested that there may be a negative variance for this factor--I did indeed see this in the TECH4. Can you give me suggestions, or may I send you my data? Thank you!

USEVARIABLES ARE ALDH2DI YEAR1 YEAR2 YEAR3 YEAR4;

MISSING IS .;
ANALYSIS: TYPE = MEANSTRUCTURE;

MODEL:

INTERCEPT BY YEAR1@1 YEAR2@1 YEAR3@1 YEAR4@1;

SLOPE BY YEAR1@0 YEAR2*.33 YEAR3*.67 YEAR4@1;

INTERCEPT ON ALDH2DI;

SLOPE ON ALDH2DI;

INTERCEPT WITH SLOPE;

OUTPUT: SAMP STAND RES MODINDICES(0) tech1 tech4;
 Linda K. Muthen posted on Thursday, September 29, 2011 - 5:38 pm
Part of the problem may be that you are not specifying the growth model correctly. If you use BY, you must fix the intercepts to zero and free the growth factor means. It would be easier if you use the growth language:

intercept slope | YEAR1@0 YEAR2*.33 YEAR3*.67 YEAR4@1;

Then everything is done correctly as the default and the default starting values are better than those for the BY statements.

I would also fit the growth model before adding ON statements.
 Lisa M. Yarnell posted on Monday, October 03, 2011 - 2:02 pm
Thanks, Linda! It worked! I have one more question: I saw both variance of latent intercept and slope, and RESIDUAL variance for latent intercept and slope reported in Bengt Muthen's writeup of a similar analysis, where he modeled the effect of predictors on latent intercept and latent slope: http://gseis.ucla.edu/faculty/muthen/Papers/Article_080.pdf

In my output, I do see the residual variances, including for latent intercept and latent slope, but not the variances of latent intercept and slope themselves.

Where do I find variance of latent intercept and latent slope in the Mplus output (not residual variances)?

Thank you so much!
 Linda K. Muthen posted on Monday, October 03, 2011 - 2:56 pm
You can find these in TECH4.
 Lisa M. Yarnell posted on Monday, October 03, 2011 - 4:13 pm
Thanks, Linda. I see this in the TECH4:

ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES

INTERCEP SLOPE ETHNIC
INTERCEP 1.677
SLOPE -0.318 1.117
ETHNIC 0.033 -0.082 0.250

But, what are the standard errors and significance levels? I saw that Bengt provided these along with the latent intercept and slope variances in his aforementioned writeup.

Thanks again.
 Lisa M. Yarnell posted on Monday, October 03, 2011 - 7:02 pm
Linda, can you tell me whether standard errors and significance levels are available for these variances?
 Bengt O. Muthen posted on Monday, October 03, 2011 - 8:25 pm
To get the SEs of thes TECH4 quantities, you would have to express the quantities in Model Constraint as new parameters. Then the SEs are obtained automcatically using the delta method.
 Lisa M. Yarnell posted on Tuesday, October 04, 2011 - 4:30 pm
Thanks, Bengt. Can you tell me what I am doing wrong here (see code below)?

Is it possible to estimate both the variance and residual variance of a latent variable using the delta method? I get the message: THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 14. THE CONDITION NUMBER IS 0.000D+00.


USEVARIABLES ARE ETHNIC YEAR1 YEAR2 YEAR3 YEAR4;

MISSING IS .;

DEFINE: ETHNIC = ETHNIC-1;

ANALYSIS: TYPE = MEANSTRUCTURE;

MODEL:

INTERCEPT SLOPE | YEAR1@0 YEAR2* YEAR3* YEAR4@1;

INTERCEPT ON ETHNIC;

SLOPE ON ETHNIC;

INTERCEPT WITH SLOPE;

MODEL CONSTRAINT:
NEW(INTERCEPT SLOPE);

OUTPUT: SAMP STAND RES MODINDICES(0) tech1 tech4;
 Bengt O. Muthen posted on Tuesday, October 04, 2011 - 9:18 pm
You have to define New parameters in Model Constraint - you haven't done that. For an example, see UG ex 5.20.
 Lisa M. Yarnell posted on Wednesday, October 05, 2011 - 9:30 pm
Bengt, I apologize for any oversight I am making, but I can't think of how to define the variances of latent intercept and latent slope using other terms in the model, except R2 + residual variance (i.e., variance explained plus variance unexplained). However, R2 and residual variance are not have explicitly named terms in the model; these are just statistics produced automatically by Mplus.

What I am looking for are the standard errors of the variances for latent intercept and latent slope in my latent growth model, as you reported in your writeup here: http://gseis.ucla.edu/faculty/muthen/Papers/Article_080.pdf

I see the variances in the TECH4 output, but not the standard errors of the variances.

Can you tell me how you would write these model constraints?
 Lisa M. Yarnell posted on Thursday, October 06, 2011 - 12:25 am
I apologize: In my previous post, I meant to write, "However, R2 and residual variance are not explicitly named terms in the model; these are just statistics produced automatically by Mplus." So, I am just not sure how to write the model constraints.
 Linda K. Muthen posted on Thursday, October 06, 2011 - 8:31 am
TECH4 does not have standard errors. For a well-fitting model, you can obtain standard errors for the variances by running the unconditional model where variances are estimated. This is probably the best solution in your case where residual variances at not model parameters.
 Lisa M. Yarnell posted on Monday, November 21, 2011 - 5:10 pm
Hello again Linda.

If a predictor GENDER with females coded as 1 affects a latent growth factor negatively, is it more appropriate to say that females showed LESS growth from the starting point (reflected by score on Initial Status) or showed SLOWER growth from the starting point?

I have heard velocity of latent growth. How does one measure velocity of latent growth? Is this something different than what is reflected by the loadings onto the latent growth factor, or the effect of a predictor such as gender on the latent growth factor?

Thank you for your knowledge,
Lisa
 Lisa M. Yarnell posted on Monday, November 21, 2011 - 9:46 pm
Linda, my quesrtion is:
How does one measure velocity of latent growth? Is this something different than what is reflected by the loadings onto the latent growth factor, or the effect of a predictor such as gender on the
latent growth factor? Is it reflected in how much freely estimated loadings onto the latent growth factor jump from one time point to another? For example, if you have several loadings: .00, *, *, *, *, 1.00 (with * meaning it is freely estimated), and the jump between loadings between time 1 and time 2 is a bigger jump than the loadings between time 4 and time 5, could you say that the first part of the trajectory has greater VELOCITY of growth? Is this velocity quantifiable?
 Bengt O. Muthen posted on Tuesday, November 22, 2011 - 5:54 pm
I am not familiar with the concept of velocity of growth - perhaps you can google for references that describe it.
 Mary Ann Simpson posted on Wednesday, November 30, 2011 - 4:24 pm
Hi, Dr. Muthen! I think Lisa M. Yarnell may be referring to the linear term in a growth model. I've heard the linear term referred to as velocity and the quadratic term as acceleration. I think the mean of the linear term in the LGM is what she's looking for.
 Lisa M. Yarnell posted on Thursday, December 15, 2011 - 11:31 am
Drs. Muthen: If in a 4 time point LGM, I set error variance to be the same across time points (which I see demonstrated in the Mplus manual), is it also possible to correlate errors for subsequent time points? Given that it is the same scale administered at multiple time points, it makes sense that scores would be correlated. However, does this introduce some sort of dependency issue in the model--either due to the errors being set to be equal across the time points, or otherwise? The model below runs well without the correlated errors, but crashes when I correlate them in either of the patterns shown below (in the correlations between YEAR1-YEAR4).

MODEL:

INTERCEPT SLOPE | YEAR1@0 YEAR2@0.385 YEAR3@0.692 YEAR4@1;

INTERCEPT ON GENDER ETHNIC;
SLOPE ON ALLELE;
INTERCEPT WITH SLOPE;

YEAR1-YEAR4 (1);

YEAR1 WITH YEAR2;
YEAR2 WITH YEAR3;
YEAR3 WITH YEAR4;

!YEAR1 WITH YEAR2 YEAR3 YEAR4;
!YEAR2 WITH YEAR3 YEAR4;
!YEAR3 WITH YEAR4;
 Linda K. Muthen posted on Thursday, December 15, 2011 - 12:55 pm
The fact that the same item is repeatedly measured is what the growth models takes into account. There may also be a need for residual covariances across time. You can look at modification indices to check this. You can hold the residual variances equal across time or not. There are generally held equal in multilevel modeling programs but that is not necessary in Mplus. If you want me to look at your output, please send it and your license number to support@statmodel.com.
 Gabriella Melis posted on Thursday, August 01, 2013 - 8:15 am
I am running a conditional LGCM with multiple indicators at each time point; each indicator is an ordered categorical. In my model I have time-varying (TVC) and time-invariant covariates (TIC). TVC and TIC are included as sets of dummies, as they are all categorical too.
The model runs well, however I loose 50% of the cases compared with the unconditional LGCM.

Then I tried to run the same conditional model after restating the names of the TIC and TVC, as we do for other models when we want Mplus to use all the available cases (full-information ML), for example:
construct1991 0n educ1 educ2;
educ1 educ2;
...

In this second case I only loose around 15% of cases, but unfortunately the model does not converge (message: number of iterations exceeded).

Please, can I do anything to avoid dropping all those cases and get my model to run?

If not, would it be correct to compare results, e.g. growth factors' estimates, between the unconditional and conditional models if the number of cases varies so much between the two?

Thank you.
 Bengt O. Muthen posted on Thursday, August 01, 2013 - 2:03 pm
Here's a trick that we suggest when you have missing on tics and the missingness for the tic at time t implies that the outcomes y at time t are missing. Score the tics at values not designated as missing data symbols. Then subjects with missing on tics will be included but not affect the likelihood due to missing on outcomes.
 ywang posted on Wednesday, March 26, 2014 - 12:56 pm
Dear Drs. Muthen,

We worked on a latent class growth model. The outcome variable is clearly very skewed. One reviewer mentioned that it has implications for erroneously identifying latent classes in a population when we treated it as a continuous variable. What can we do for the skewed outcome variable in latent class growth modeling (and the parallel latent growth modeling linked to another outcome)?
Thanks a lot!
 Linda K. Muthen posted on Thursday, March 27, 2014 - 2:02 pm
Two issues are relevant in this situation. One is whether the classes make substantive sense based on theory. The other is whether external validity can be demonstrated using a distal outcome. See the following paper which is available on the website:

Muthén, B. (2003). Statistical and substantive checking in growth mixture modeling. Psychological Methods, 8, 369-377.
 Holly Andrewes posted on Wednesday, February 11, 2015 - 2:10 pm
Dear Drs. Muthens,

I have developed a quadratic LGCM with 5 time points (varying times of observation).

I notice that when I add a time invariant covariate (dichotomous) to the model, it will not converge, stating that there is a "non-positive definite fisher information matrix". The output also states: PROBLEM INVOLVING PARAMETER 9. Using TECH 1 I see that this parameter is the correlation between my covariate and the intercept. By chance I have managed to resolve the problem by defining time differently (time = time/24).

1. However, I am unclear as to what the initial problem was. Are you able to tell me what this might have been?

2. Also, unfortunately, this new time scale does not make much sense in my model, is there any other way of resolving this problem without changing the time scale?

Thank you very much for your help!
 Bengt O. Muthen posted on Wednesday, February 11, 2015 - 6:10 pm
Since you mention a correlation, it sounds like you have both

i ON x;

and

i WITH x;

which would not be identified. But maybe I am not understanding your situation, in which case you should send the problematic output to support along with your license number.
 Holly Andrewes posted on Monday, February 16, 2015 - 11:54 pm
Dear Bengt,

Thank you very much for your reply. I believe I have solved this problem. However, I have a remaining issue, which is that the Log-likelhood of the unrestricted model (H1) appears to be more negative than the restricted model (H0 - my model). This appears to imply that the my model fits the data better than this unrestricted model, which doesn't seem to make sense. This also provides a negative chi-square when comparing the model. I am wondering if you have any idea why this might be the case, and what I might be able to do to rectify this?

Thank you very much for all your help.
 Bengt O. Muthen posted on Tuesday, February 17, 2015 - 8:52 am
Please send this output to Support along with your license number.
 Kimmo Sorjonen posted on Tuesday, March 10, 2015 - 3:15 am
Dear Linda or Bengt!
I have run a latent growth model where I regress the intercept and slope growth factors on a continuous predictor X. X has a negative effect on the intercept and a positive effect on the slope while the correlation between the intercept and the slope is negative. The suspicion arises that the positive effect of X on the slope might be confounded by the negative association between X and the intercept (those high in X tend to have lower starting values and therefore more room to increase in the outcome). Would it make sense to let the slope regress on both X and the intercept? Would this give me the effect of X on the slope while controlling for the intercept?
 Bengt O. Muthen posted on Tuesday, March 10, 2015 - 4:35 pm
That's a perfectly fine approach.
 Kelly M Allred posted on Wednesday, December 16, 2015 - 11:22 am
I am trying to estimate the unconditional model for a piecewise latent growth curve model using the following syntax:

Alpha_C by Crit_1@1 Crit_2@1 Crit_3@1 Crit_4@1 Crit_5@1;
Beta1_C by Crit_1@0 Crit_2@1 Crit_3@2 Crit_4@2 Crit_5@2;
Beta2_C by Crit_1@0 Crit_2@0 Crit_3@0 Crit_4@1 Crit_5@2;
[Crit_1@0 Crit_2@0 Crit_3@0 Crit_4@0 Crit_5@0];
[Alpha_C Beta1_C Beta2_C];
Alpha_C with Beta1_C Beta2_C;
Beta1_C with Beta2_C;

I am getting the following message:
THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE...PROBLEM INVOLVING VARIABLE BETA2_C.

I have looked at the regular and TECH4 output and am unable to figure out why I can getting this message. Any ideas on how to remedy this problem?
 Linda K. Muthen posted on Wednesday, December 16, 2015 - 11:49 am
Please send the output and your license number to support@statmodel.com.
 Kelly M Allred posted on Wednesday, July 06, 2016 - 8:42 am
I am trying to estimate a conditional model for a piecewise latent growth curve model using the following truncated syntax which does not show the syntax for estimating the latent intercepts (alphas) and slopes (betas):

Alpha_P with Beta1_P Beta2_P;
Beta1_P with Beta2_P;

Beta1_P Beta2_P on Alpha_C;
Beta1_C with Beta2_C;

Alpha_C with Beta1_C Beta2_C;

Beta1_P with Beta1_C;
Beta2_P with Beta2_C;
Panic_M with Alpha_P Beta1_P Beta2_P;
Crit_M with Alpha_C Beta1_C Beta2_C;

Panic_7@0;
Panic_M with Crit_M;

I am receiving the following message:
THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE...PROBLEM INVOLVING VARIABLE BETA2_C.

I have checked the regular and tech outputs as well as correlations and variances/res. variances. Any ideas on how to diagnose the problem?
 Kelly M Allred posted on Wednesday, July 06, 2016 - 8:43 am
With regard to the model in the previous post, I also receive this message:

THE STANDARD ERRORS FOR H1 ESTIMATED SAMPLE STATISTICS COULD NOT BE COMPUTED. THIS MAY BE DUE TO LOW COVARIANCE COVERAGE. THE ROBUST CHI-SQUARE COULD NOT BE COMPUTED.

Does this message mean that parameter estimates for this model should not be trusted? Also, how does one determine model fit if the chi-square cannot be computed?
 Linda K. Muthen posted on Wednesday, July 06, 2016 - 3:14 pm
Please send the output and your license number to support@statmodel.com.
 Thula Chelvan posted on Wednesday, January 03, 2018 - 11:44 am
I'm running a 2-group latent growth model of longitudinal model to find support for invariance (the slope loadings and residual variances at each time point) between the two treatment groups in my sample. I want to justify collapsing both groups into one sample to run subsequent analysis.

When I first ran the two groups together, the model of random linear and random quadratic slope fit well, but the variance of each growth factor was not significant..

When running the model as a 2-group, the variance of each growth factor within the each group is now significant!

Can this be due to increased power with a 2-group approach?
 Bengt O. Muthen posted on Wednesday, January 03, 2018 - 2:34 pm
Check that each group in the 2-grp run has the same number of parameters as each group's run. Perhaps invariance of some parameters is defaulted in which case more power is obtained.
 Thula Chelvan posted on Thursday, January 04, 2018 - 8:11 am
Thank you for your response.

I have a follow-up question. If variance of a growth parameter is found to be initially not significant,

(1) are you justified in adding a covariate (i.e. treatment arm) to the model?

(2) Given the above results, what does it mean if the effect of the covariate on the growth factor is found to be significant?

Many thanks
 Bengt O. Muthen posted on Thursday, January 04, 2018 - 4:28 pm
(1) Yes, because you have a different amount of power to detect the variation when a covariate is included.

(2) It means that there is significant variation in the growth factor.
 Marc Goodrich posted on Thursday, April 05, 2018 - 9:33 am
Hi Dr. Muthen,

I am running a quadratic latent growth model with four time points, and the intercept fixed at time 1. The unconditional model yieled a non-significant negative residual variance for the linear slope term, so this term was fixed to 0. When various predictors of growth parameters are added in the conditional model, I am receiving an error that there is a non-positive definite first-order derivative product matrix, and the error indicates that the problematic parameter is the variance of the latent intercept in the psi matrix. I have tried a number of solutions, including fixing the residual variance, changing the starting values, etc., but nothing seems to solve the problem. An additional issue seems to be that the conditional model has worse fit than the unconditional model

Do you have any suggestions for how to proceed? Are the estimates for prediction of the growth parameters from the conditional model interpretable?

Thank you.
 Bengt O. Muthen posted on Thursday, April 05, 2018 - 2:28 pm
Perhaps you have a binary covariate that has its mean or variance mentioned in which case the message can be ignored. If not, send output to Support along with license number.

The conditional model imposes more restrictions and therefore has more ways of misspecification.
 Marc Goodrich posted on Thursday, April 05, 2018 - 3:48 pm
Thanks for the quick response! There are not any binary covariates, but there are some Likert-type covariates that are positively or negatively skewed for which the means are estimated. Could this result in the same situation as the binary covariate?
 Bengt O. Muthen posted on Thursday, April 05, 2018 - 5:52 pm
No. You should send your output to Support along with your license number so we can diagnose.
 Thula_Chelvan posted on Monday, December 03, 2018 - 4:52 am
Hello,

I'm wondering if someone can comment on how to interpret i and s in a conditional LGCM after a predictor has been included? Do you still look at the significance of these intercepts, and what does that mean?

Thanks
 Bengt O. Muthen posted on Monday, December 03, 2018 - 4:07 pm
No, the intercepts of the i and s growth factors are typically not of interest.
 Thomas Kirsh posted on Monday, May 20, 2019 - 3:06 pm
Hello,

I'm running three different latent curve growth models using linear, quadratic, and cubic trajectories, over 8 time points. To compare the models, my procedure is

1. Compare the VLMP and LRT p-values for each class in the trajectory. If, for example, 3 classes are more significant than 4 classes in the linear model, then I say that the linear model found 3 classes to be significant.

2. Find the BIC for the trajectory that the model found significant for each model (linear, quadratic, cubic). Compare the BIC values and the model with the BIC closest to zero is the model that fits the data the best (for example, if the BIC for the linear model with 4 classes is 11344 and the BIC for the quadratic model with 3 classes is 12553, then the linear model fits the data best).

Is this the best or most accepted procedure for comparing models that may select different numbers of significant classes? If so, is there a citation I can use?

Thanks,
Tom
 Bengt O. Muthen posted on Monday, May 20, 2019 - 5:33 pm
I would simply use only BIC.
 Thomas Kirsh posted on Tuesday, May 21, 2019 - 7:04 am
Thanks for your help, Dr. Muthen. To clarify, do you mean use only BIC to decide the significant classes (step 1) or use only BIC in step 2 to determine best model fit?
 Bengt O. Muthen posted on Tuesday, May 21, 2019 - 5:25 pm
Use only BIC.
 May Gong posted on Monday, July 29, 2019 - 6:50 am
Dear Dr. Muthen,

We are doing the conditional latent growth model in which examining the prediction of X. However, the model warns that "The latent variable covariance matrix (PSI) is not positive definite. This could indicate a negative variance/residual variance for a latent variable, a correlation greater or equal to one between two latent variables, or a linear dependency among more than two latent variables. Problem involving variable s."

And then, we can found that the intercept and residual variances of s is negative. I wonder how can fix this problem. Thanks in advance!

Best wishes,
May Gong
2019/7/29
 Bengt O. Muthen posted on Monday, July 29, 2019 - 5:23 pm
We need to see your full output - send to Support along with your license number.
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