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 1010
#define max(a,b) ((a<b)?b:a)

char s[MAX], t[MAX];
int dp[MAX][MAX],path[MAX][MAX];

int LCS() {
  int n = strlen(s);
  int m = strlen(t);
  memset(dp,0,sizeof dp);
  int i,j,maxi=0;
  for(i=0;i<n;i++) {
    for(j=0;j<m;j++) {
      if( i ) dp[i][j] = max(dp[i][j],dp[i-1][j]);
      if( j ) dp[i][j] = max(dp[i][j],dp[i][j-1]);
      if( s[i] == t[j] ) {
	dp[i+1][j+1] = max(dp[i+1][j+1],dp[i][j]+1);
	maxi = max(maxi,dp[i+1][j+1]);
      }
    }
  }
  return maxi;
}

int main() {
  int n,i;
  scanf("%d",&n);
  for(i=0;i<n;i++) {
  scanf(" %s",&s[0]);
  scanf(" %s",&t[0]);
  printf("%d\n",LCS());
  }
  return 0;
}