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<stdlib.h>
#include<string.h>


int LCSLength(char x[], char y[]) {
  int i, j;
  int xLen, yLen;
  int** c;  

  xLen = strlen(x);
  yLen = strlen(y);

  c = (int**)malloc( sizeof(int*) * (xLen+1) );
  for( i=0; i<=xLen; i++) {
    c[i] = (int*)malloc( sizeof(int) * (yLen+1) );
  }

  for( i=1; i<=xLen; i++) {
    c[i][0] = 0;
  }
  for( j=0; j<=yLen; j++) {
    c[0][j] = 0;
  }

  for( i=1; i<=xLen; i++) {
    for( j=1; j<=yLen; j++) {
      if( x[i-1] == y[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];
      }

    }
  }

  j = c[xLen][yLen];

  for( i=0; i<=xLen; i++) {
    free(c[i]);
  }
  free(c);

  return j;

}

main() {
  
  int n, i;
  char x[1001], y[1001];

  scanf("%d", &n);

  for( i=0; i<n; i++) { 
    scanf("%s%s", x, y);
    
    printf("%d\n", LCSLength( x, y) );
	   
  }

  return 0;
  
}