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 1007

char x[N],y[N];
int c[N][N];

int L[N];

int m,n;

int LCSlength(char *,char *);

int  main(){
  int i,j,n;

  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%s",x+1);
    scanf("%s",y+1);
    L[i] =  LCSlength(x,y);
  }
  for(i=0;i<n;i++){
     printf("%d\n",L[i]);
  }
  return 0;
}



int LCSlength(char *x,char *y){
  int m,n;
  int i,j;
  int max=0;

  m = strlen(x+1);
  n = strlen(y+1);
  
  for(i=0;i<=m;i++) c[i][0] = 0;
  for(j=0;j<=n;j++) c[0][j] = 0;

  for(i=1;i<=m;i++){
    for(j=1;j<=n;j++){
      
      if(x[i] == y[j]){
	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];
      if(max < c[i][j]) max = c[i][j];      
    }
  }

  return max;

}