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<math.h>
#include<string.h>
#define N 1000

int LCS(char X[], char Y[]){
  int m, n, max, i, j;
  int c[N+1][N+1];

  m = strlen(X);
  n = strlen(Y);
  max = 0;

  X[m+2] = '\0';
  for( i = m+1 ; i > 0 ; i-- ){
    X[i] = X[i-1];
  }
  X[0] = ' ';

  Y[n+2] = '\0';
  for( i = n+1 ; i > 0 ; i-- ){
    Y[i] = Y[i-1];
  }
  Y[0] = ' ';

  for( i = 0 ; i <= m ; i++ ) c[i][0] = 0;
  for( j = 0 ; j <= n ; j++ ) c[0][j] = 0;


  for( i = 1 ; i <= m ; i++ ){
    for( j = 1 ; j <= n ; j++ ){
      if( X[i] == Y[j] ) c[i][j] = c[i-1][j-1] + 1;
      else {
        c[i][j] = fmax(c[i-1][j], c[i][j-1]);
      }
      max = fmax(max, c[i][j]);
    }
  }
  return max;
}

int main(){
  char str1[N], str2[N];
  int n, i;

  scanf("%d",&n);

  for( i = 0 ; i < n ; i++ ){
    scanf("%s%s",str1,str2);
    printf("%d\n",LCS(str1,str2));
  }
  return 0;
}