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 1005
  
int main()
{ 
  int tum, mum, num;
  int i, j, c[MAX][MAX];
  char a[MAX], b[MAX];
  
  scanf("%d", &tum);
  
  while(tum--){
    scanf("%s", a);
    scanf("%s", b);
    mum = strlen(a);
    num = strlen(b);
    for(i=1; i<=mum; i++){
      c[i][0] = 0;
    }
    for(j=0; j<=num; j++){
      c[0][j] = 0;
    }
    for(i=1; i<=mum; i++){
      for(j=1; j<=num; j++){
        if(a[i-1] == b[j-1]){
          c[i][j] = c[i-1][j-1] + 1;
        }
        else{
          if(c[i-1][j] >= c[i][j-1]) c[i][j] = c[i-1][j];
          else c[i][j] = c[i][j-1];
        }
      }
    }
    printf("%d", c[mum][num]);
    printf("\n");
  }
 
  return 0;
}