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>
void LCSLength(void);

char X[1001], Y[1001];

int c[1001][1001];

int n, m;

main(){
  int i, q;

  scanf("%d",&q);

  for(i = 0; i < q; i++){

    scanf("%s", X);

    scanf("%s", Y);

    LCSLength();

    printf("%d\n", c[m][n]);
  }

  return 0;
}

void LCSLength(void){

  int i, j;

  m = strlen(X);

  n = strlen(Y);

  for(i = 1; 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-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];
    }
  }
}