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 N (1000 + 10)

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

int A[N][N];
char a[N], b[N];

int LCS(void);

int main(void){
  int n, i;
  scanf("%d\n", &n);
  for ( i = 0; i < n; i++ ){
    fgets(a, N, stdin);
    fgets(b, N, stdin);
    printf("%d\n", LCS());
  }
  return 0;
}

int LCS(void){
  const int a_size = strlen(a) - 1;
  const int b_size = strlen(b) - 1;
  int i, j;

  memset(A, 0, sizeof(A));

  for (i = 1; i < a_size + 1; ++i) {
    for (j = 1; j < b_size + 1; ++j) {
      int mx = max(A[i][j - 1], A[i - 1][j]);
      if (a[i - 1] == b[j - 1]) {
        mx = max(mx, A[i - 1][j - 1] + 1);
      }
      A[i][j] = mx;
    }
  }

  return A[a_size][b_size];
}