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 <stdlib.h>
#include <string.h>

#define N 1010

void LCSLength();

char X[N], Y[N];
int Xlen, Ylen;

int main(){
  int num, i, j;
  char c;
  
  scanf("%d%*[^\n]", &num);
  getchar();
  
  for(i = 0; i < num; i++){
    strcpy(X, "");
    strcpy(Y, "");

    j = 0;
    while((c = getchar()) != '\n'){
      X[j] = c;
      j++;
    }
    X[j] = '\0';
    Xlen = j;
    fflush(stdin);
    
    j = 0;
    while((c = getchar()) != '\n'){
      Y[j] = c;
      j++;
    }
    Y[j] = '\0';
    fflush(stdin);
    
    Ylen = j;

    LCSLength();
  }
  /*
  for(i = 0; i < num; i++){
    printf("%d\n", C[i]);
  }
  */
  
  return 0;
}

void LCSLength(){
 long long int c[N][N];
  int i, j;

  // printf("X = %s, Xlen = %d\n", X, Xlen);
  //printf("Y = %s, Ylen = %d\n", Y, Ylen);
  
  for(i = 1; i <= Xlen; i++){
    c[i][0] = 0;
  }
  for(j = 1; j <= Ylen; j++){
    c[0][j] = 0;
  }

  for(i = 1; i <= Xlen; i++){
    for(j = 1; j <= Ylen; 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];
	}
      }
    }
  }
  /*
  for(i = 0; i <= Xlen; i++){
    for(j = 0; j <= Ylen; j++){
      printf(" %2d", c[i][j]);
    }
    printf("\n");
  }
  */
  printf("%d\n", c[i-1][j-1]);
 
}