Problem:
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Longest Common Subsequence

For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \{a,b,c,b,d,a,b\}$ and $Y = \{b,d,c,a,b,a\}$, the sequence $\{b,c,a\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\{b,c,a\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\{b,c,b,a\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\{b,c,b,a\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.

Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Input

The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.

Output

For each dataset, print the length of LCS of $X$ and $Y$ in a line.

Constraints

$1 \leq q \leq 150$

$1 \leq$ length of $X$ and $Y$ $\leq 1,000$

$q \leq 20$ if the dataset includes a sequence whose length is more than 100

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

char X[1001];
char Y[1001];



//int i,j;

/*void print(int k){
    
    printf("%d\n",k);
    
}*/



/*int lcs(int i,int j){
    int len1,len2;
    
  //  i=strlen(X)-1;
  //  j=strlen(Y)-1;
    
    if(i<0||j<0)return 0;
    
    if(L[i][j] !=-1)return(L[i][j]);
    
    if(X[i]==Y[j]){
        L[i][j]=lcs(i-1,j-1)+1;
        return L[i][j];
    }else{
        len1=lcs(i-1,j);
        len2=lcs(i,j-1);
        if(len1>=len2)L[i][j]=len1;
         else  L[i][j]=len2;
        return L[i][j];
    }

    
}*/


void lcs(int m,int n){
    int i,j;
    int L[1001][1001];
    
    for(i=0;i<=m;i++){
        for(j=0;j<=n;j++){
            L[i][j]=0;
        }
    }
    
    
    
    
    for(i=1;i<=m;i++){
        for(j=1;j<=n;j++){
            if(X[i-1]==Y[j-1]){
                L[i][j]=L[i-1][j-1]+1;
            }
            else if(L[i-1][j]>=L[i][j-1])L[i][j]=L[i-1][j];
            else L[i][j]=L[i][j-1];
        }
    }
    
    printf("%d\n",L[m][n]);
}



int main(void){
    int s,p,set;
    int i,j;
    
    
  /*  for(i=0;i<100;i++){
        for(j=0;j<100;j++){
            L[i][j]=0;
        }
    }
    */
    
    
    scanf("%d",&set);
    
    for(i=0;i<set;i++){
    scanf("%s",X);
    scanf("%s",Y);
    lcs(strlen(X),strlen(Y));
    //printf("%d",L[i][j]);
       // printf("%d\n",lcs(/*m-1,n-1*/));
       
    }
    
    return 0;
    
    
}