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 MAX 1001

typedef struct{
  char X[MAX];
  char Y[MAX];
}Shuto;


int main()
{
  int n,xl,yl,i,j,k,c[MAX][MAX];

  scanf("%d",&n);
  if(n < 1 || n > 150)exit(1);

  Shuto kumi[n];

  for(i = 0 ; i < n ; i++)
    {

      scanf("%s",&kumi[i].X);
      scanf("%s",&kumi[i].Y);
    }

  for(i = 0 ; i < n ; i++)
    {

      xl = strlen(kumi[i].X);
      yl = strlen(kumi[i].Y);

      c[0][0] = 0;

      for(j = 1 ; j <= xl ; j++)c[j][0] = 0;
      for(j = 1 ; j <= yl ; j++)c[0][j] = 0;

      for(j = 1 ; j <= xl ; j++)
	{
	  for(k = 1 ; k <= yl ; k++)
	    {
	      if(kumi[i].X[j-1] == kumi[i].Y[k-1])
		{
		  c[j][k] = c[j-1][k-1] + 1;
		}
	      else if(c[j-1][k] >= c[j][k-1])
		{
		  c[j][k] = c[j-1][k];
		}
	      else
		{
		  c[j][k] = c[j][k-1];
		}
	    }
	}

      printf("%d\n",c[xl][yl]);
    }

  return 0;
}