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>

#define N 2000
void LCS(char[],char[]);

int main(int argc,char *argv[]){
  int i,n;
  char a[N],b[N];

  scanf("%d",&n);
  for(i=0; i<n; i++)
    {
      scanf("%s",a);
      scanf("%s",b);
      LCS(a,b);


    }
  return 0;



}
void LCS(char A[], char B[])
{
  int i,j,m,n;

    m = strlen(A);
    n = strlen(B);
    int a[m+1][n+1];

    for(i=0; i<=m; i++){
      for(j=0; j<=n; j++){
        a[i][j] = 0;
      }
    }

    for(i = 1; i <= m; i++){
      for(j = 1; j <= n; j++){
    if(A[i-1] == B[j-1])
      a[i][j] = a[i-1][j-1] + 1;
    else if(a[i-1][j] >= a[i][j-1])
      a[i][j] = a[i-1][j];
    else a[i][j] = a[i][j-1];
      }
    }
    printf("%d\n",a[m][n]);

}