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<string.h>
#define N 1000
#define M 150

int LCSLength(char *, int, int, char *, int, int);
int max(int, int);

int main()
{
  int q, i, r[M];
  char X[N], Y[N];

  scanf("%d", &q);

  for(i = 0; i < q; i++){
    scanf("%s", X);
    scanf("%s", Y);

    r[i] = LCSLength(X, 0, (int)strlen(X), Y, 0, (int)strlen(Y));
  }

  for(i = 0; i < q; i++){
    printf("%d\n", r[i]);
  }
  
  return 0;
}

int LCSLength(char *X, int i, int size_X, char *Y, int j, int size_Y)
{
  
  // 終了条件
  if(i >= size_X || j >= size_Y) return 0;
  
  // 一致する場合は両配列のインデックスを +1 だけ進める
  else if(X[i] == Y[j]) return 1 + LCSLength(X, i + 1, size_X, Y, j + 1, size_Y);
  
  // 一致しない場合は部分問題として扱う
  else return max(LCSLength(X, i + 1, size_X, Y, j, size_Y), LCSLength(X, i, size_X, Y, j + 1, size_Y));
  
}

int max(int a, int b)
{
  if (a >= b) return a;
  else return b;
}