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 <stdlib.h>
#include <string.h>
 
char len1[1001], len2[1001];
 
LCS(int x, int y){
  int i,j,c[1001][1001];
 
   for(i=0;i<=x;i++){
      c[i][0]=0;
  }
 
  for(j=0;j<=y;j++){
    c[0][j]=0;
  }
 
  for(i=1;i<=x;i++){
    for(j=1;j<=y;j++){
      if(len1[i-1]==len2[j-1]){
    c[i][j]=c[i-1][j-1]+1;
      }else if(c[i-1][j-1]>=c[i][j-1]){
    c[i][j] = c[i-1][j];
      }else{
    c[i][j] = c[i][j-1];
      }
    }
  }
 
  if(i==0 || j==0){
    return 0;
  }else if(i>0 && j>0 && x == y){
    return c[i-1][j-1];
  }else if(i>0 && j>0 && x!=y){
    return c[x][y];
  }
}
 
main(){
  int n,i,l1,l2,result;
 
  scanf("%d", &n);
   
  for(i=0;i<n;i++){
    scanf("%s",len1);
    scanf("%s",len2);
    l1=strlen(len1);
    l2=strlen(len2);
    
    result=LCS(l1,l2);
    printf("%d\n",result);
  }
  return 0;
}