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>
typedef struct{
  char X[1001];
  char Y[1001];
  int xlen;
  int ylen;
}Q;

main(){
  int a,b,i,j,k,q,xlen,ylen,z[1001][1001];
  Q n[300];
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%s",n[i].X);
    scanf("%s",n[i].Y);
    n[i].xlen = strlen(n[i].X);
    n[i].ylen = strlen(n[i].Y);
  }
  for(i=0;i<q;i++){
    for(j=1;j<=n[i].xlen;j++)z[j][0]=0;
    for(k=1;k<=n[i].ylen;k++)z[0][k]=0;

    for(a=1;a<=n[i].xlen;a++){
      for(b=1;b<=n[i].ylen;b++){
        if(n[i].X[a-1] == n[i].Y[b-1])
          z[a][b] = z[a-1][b-1] + 1;
        else if(z[a-1][b]>=z[a][b-1])
          z[a][b] = z[a-1][b];
        else
          z[a][b] = z[a][b-1];
      }}
    printf("%d\n",z[n[i].xlen][n[i].ylen]);
  }

return 0;
}