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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main(){
  int n;
  char x[1001],y[1001];
  int dp[1002][1002];
  
  //initialize
  for(int i=0;i<1002;i++){
    for(int j=0;j<1002;j++){
      dp[i][j]=0; 
    }
  }

  //load n;
  cin >>n;
  for(int i=0;i<n;i++){
    //load string
    scanf("%s",x);
    scanf("%s",y);
    //dp
    for(int j=0;j<strlen(x);j++){
      for(int k=0;k<strlen(y);k++){
	if(x[j]==y[k]){
	  dp[j+1][k+1]=dp[j][k]+1;
	}

	else{
	  if(dp[j][k+1]>dp[j+1][k]){
	    dp[j+1][k+1]=dp[j][k+1];
	  }

	  else{
	    dp[j+1][k+1]=dp[j+1][k];
	  }
	}
      }
    }

    //print
    cout <<dp[strlen(x)][strlen(y)]<<endl;
  }



  return 0;
}