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 o, i, j, k;
  
  int alen, blen;
  
  char a[1000],b[1000];
  
  int c[1001][1001];
  
  scanf("%d\n",&o);

  
  for (k=0;k<o;k++) {
    
    scanf("%s",a);
    alen = strlen(a);
    
    if (a[alen-1]=='\n'){
      a[alen-- -1]='\0';
    }
    scanf("%s",b); 
    blen = strlen(b);
    
    if (b[blen-1]=='\n'){
      
      b[(blen--)-1]='\0';
      
    }
    
    c[0][0]=0;
    
    for (i=0; i<alen; i++){
      
      for (j=0; j<blen; j++){
	
        if (a[i]==b[j]){
	  
          c[i+1][j+1] = c[i][j]+1;
	  
	}
	
        else
	  {
	    if(c[i][j+1] > c[i+1][j])  c[i+1][j+1] = c[i][j+1];
	    else c[i+1][j+1] = c[i+1][j];
	   
	  }
      }
    }
    
    printf("%d\n",c[alen][blen]);
    
  }
  return 0;
}