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>
 int main(){
 char x[1000],y[1000];
 int n,i,j;
 scanf("%d",&n);
 for(i = 0;i < n;i++){
    scanf("%s %s",x, y);
   printf("%d\n",cal(x,y));
 }
 return 0;     
}
  
int cal(char x[], char y[]){
  
 int i,j;
 int m = strlen(x);
 int n = strlen(y);
 int cou[m+1][n+1];
  
 for(i = 0;i <= m; i++){
  cou[i][0] = 0;
  }
 for(j = 0;j <= n; j++){
  cou[0][j] = 0;
  }
 for(i = 1;i <= m; i++){
  for(j = 1;j <= n; j++){
    if(x[i-1] == y[j-1])
     cou[i][j] = cou[i-1][j-1] + 1;
    else if(cou[i-1][j] >= cou[i][j-1]){
     cou[i][j] = cou[i-1][j];
  }
    else
     cou[i][j] = cou[i][j-1];
  }
 }
 return cou[m][n];
}