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 max(int,int);
int LCSlength(char* ,char *);
int main(){
  int n,i;
  char X[1001],Y[1001];
  scanf("%d",&n);
  for(i = 0;i < n;i++){
    scanf("%s%s",X,Y);

    printf("%d\n",LCSlength(X,Y));
  }
  return 0;
}

int max(int a, int b){ return a < b ? b : a;}

int LCSlength(char *X,char *Y){
  int xlen,ylen,i,j;
  int Z[1001][1001];
  xlen = strlen(X);
  ylen = strlen(Y);
  for(i = 0; i < xlen+1;i++){
    Z[i][0] = 0;
  }
 for(i = 0; i < ylen+1;i++){
    Z[0][i] = 0;
  }



  for(i = 1;i <= xlen; i++){
    for(j = 1; j <= ylen; j++){
      if(X[i-1] == Y[j-1]) Z[i][j] = Z[i-1][j-1] + 1;
      else Z[i][j] = max(Z[i-1][j], Z[i][j-1]);
    }
  }

  

  return Z[xlen][ylen];
}