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 1000
#define max(a,b) (a < b ? b : a)

int main(){
  int Q,i,j,l1,l2,dp[MAX+1][MAX+1];
  char X[MAX+1],Y[MAX+1];
  scanf("%d",&Q);
  while(Q--){
    scanf("%s %s",X,Y);
    l1 = strlen(X);
    l2 = strlen(Y);
    memset(dp,0,sizeof(dp));
    for(i = 0 ; i < l1 ; i++){
      for(j = 0 ; j < l2 ; j++){
        if(X[i] == Y[j]){
          dp[i+1][j+1] = dp[i][j] + 1;
        }else{
          dp[i+1][j+1] = max(dp[i+1][j],dp[i][j+1]);
        }
      }
    }
    printf("%d\n",dp[l1][l2]);
  }
  return 0;
}