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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

int solve(char *a, char *b, int i, int j) {
  if(a[i] == '\0' || b[j] == '\0') return 0;
  if(a[i] == b[j]) return solve(a,b,i+1,j+1)+1;
  int k;
  for (k = i+1; a[k] != '\0'; k++) if(a[k] == b[j]) break;
  int v1 = solve(a,b,k,j);
  for (k = j+1; b[k] != '\0'; k++) if(a[i] == b[k]) break;
  int v2 = solve(a,b,i,k);
  int v3 = solve(a,b,i+1,j+1);
  if(v3 < v1) return (v1 > v2)? v1: v2;
  return (v3 > v2)? v3: v2;
}

int main()
{
  int i,q;
  char *a,*b;
  scanf("%d",&q);
  for (i = 0; i < q; i++) {
    a = (char*)malloc(sizeof(char) * 1000);
    b = (char*)malloc(sizeof(char) * 1000);
    scanf("%s %s",a,b);
    printf("%d\n",solve(a,b,0,0));
  }
  return 0;
}