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>

int data[1001][1001];
char set1[1000], set2[1000];
int shori(int, int);
void shokika();

int main(){
  
  int n, i, x, y, max, temp; /*x:横 y:縦*/

  for(i = 0; i <= 1000; i++)
    {
      data[0][i] = 0;
      data[i][0] = 0;
    }

  scanf("%d",&n);
  
  for(i = 0; i < n; i++)
    {
      shokika();
      scanf("%s%s",set1,set2);
      for(y = 0; y <= 1000; y++)
	{
	  //printf("set2[%d] = %c\n",y,set2[y]);
	  if(set2[y] == NULL)break;
	  for(x = 0; x <= 1000/*set1[x] != NULL*/; x++)
	    {
	      //printf("set1[%d] = %c\n",x,set1[x]);
	      if(set1[x] == NULL)break;
	      temp = shori(x + 1, y + 1);
	      if(temp > max || x == 0)max = temp;
	    }
	}
      printf("%d\n",max);
    }
  
  return 0;
}

int shori(int x, int y)
{
  int n, flag = 0;
  
  if(set1[x - 1] == set2[y - 1])flag = 1;
  
  switch(flag)
    {
    case 0:
      
      if(data[x-1][y] > data[x][y-1])return data[x][y] = data[x-1][y];
      
      else return data[x][y] = data[x][y-1];
      
    case 1:
      
      return data[x][y] = data[x-1][y-1] + 1;
    }
}

void shokika()
{
  int i;
  for(i = 0; i <= 1000 && set1[i] != NULL; i++)set1[i] = NULL;
  for(i = 0; i <= 1000 && set2[i] != NULL; i++)set2[i] = NULL;
}