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: Wrong Answer)

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

int main(){
  int i, j, k, n, z[1000][1000], l1, l2;
  char s1[1000], s2[1000];
  for(i = 0; i < l1; i++)
    z[i][0] = 0;
  for(j = 0; j < l2; j++)
    z[0][k] = 0;
  scanf("%d",&n);
  for(i = 0; i < n; i++){
    scanf("%s",s1);
    l1 = strlen(s1);
    scanf("%s",s2);
    l2 = strlen(s2);
    for(j = 1; s1[j-1]!='\0'; j++){
      for(k = 1; s2[k-1] != '\0'; k++){
	if(s1[j-1] == s2[k-1]) {
	  z[j][k] = z[j-1][k-1] + 1;
	  //printf("%d",z[j][k]);
	}
	else if(z[j-1][k] >= z[j][k-1]){
	  z[j][k] = z[j-1][k];
	  //printf("%d",z[j][k]);
	}
	else {
	  z[j][k] = z[j][k-1];
	  //printf("%d",z[j][k]);
	}
      }
    }
    printf("%d\n",z[j-1][k-1]);
  }
  return 0;
}