Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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)

/*
  Longest Common Subsequence problem
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_LEN 1001

int max(int, int);
int LCSLength(char x[], char y[], int, int);

int main()
{
  int i, q, *res;
  char x[MAX_LEN], y[MAX_LEN];

  scanf("%d", &q);
  res = (int *)malloc( sizeof(int) * q );

  for( i = 0; i<q; i++ )
    {
      scanf("%s%s", x, y);
      res[i] = LCSLength(x, y, strlen(x), strlen(y));          /*　最長共通部分列の長さを求める　*/
    }

  /*　各データセットの最長共通部分列の長さを出力　*/
  for( i = 0; i<q; i++ )
    printf("%d\n", res[i]);
  
  return 0;
}


/* int型整数 a,b　の大きい方を求めて返す関数　*/
int max(int a, int b)
{
  if( a >= b )
    return a;
  else
    return b;
}


/*　文字列xとyの最長共通部分列の長さを求めて返す関数　*/
int LCSLength(char x[], char y[], int x_len, int y_len)
{
  int i, j, c[x_len+1][y_len+1], max_len = 0;

  /*　cの0行目と0列目を０で初期化　*/
  for( i = 1; i<=x_len; i++ )
    c[i][0] = 0;
  for( j = 0; j<=y_len; j++ )
    c[0][j] = 0;

  /* c[i][j]を求める　*/
  for( i = 1; i<=x_len; i++ )
    {
      for( j = 1; j<=y_len; 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];

	  max_len = max(max_len, c[i][j]);
	}
    }
  
  return max_len;
}