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()
{
  int q,a,b;
  char S1[1001],S2[1001];
  
  scanf("%d",&q);
  for(a=0;a<q;a++){
    scanf("%s",S1);
    scanf("%s",S2);
    printf("%d\n",LCS(S1,S2));
  }
  
  return 0;
}

int LCS(char *S1,char *S2)
{
  int c[1001][1001],a,b,m,n;
  m=strlen(S1)+1;
  n=strlen(S2)+1;
  
  for(a=1;a<m;a++)c[0][a]=0;
  for(a=0;a<n;a++)c[a][0]=0;
  
  for(a=1;a<m;a++){
    for(b=1;b<n;b++){
      if(S1[a-1]==S2[b-1])c[a][b]=c[a-1][b-1]+1;
      else if(c[a-1][b]>=c[a][b-1])c[a][b]=c[a-1][b];
      else c[a][b]=c[a][b-1];
    }
  }
  
  return c[m-1][n-1];
}