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 main()
{
  int n, i, j, k, xlen, ylen;
  char x[1003], y[1003];
  int c[1001][1001];
  char buf[20];
  buf[19]='\0';
  scanf("%d\n", &n);
  for (k=0; k<n; k++) {
    fgets(x, 1002, stdin);
    xlen = strlen(x);
    if (x[xlen-1]=='\n') x[xlen-- -1]='\0';
    fgets(y, 1002, stdin);
    ylen = strlen(y);
    if (y[ylen-1]=='\n') y[ylen-- -1]='\0';

    c[0][0]=0;
    
    for (i=0; i<xlen; i++) {
      for (j=0; j<ylen; j++) {
        if (x[i]==y[j]) 
          c[i+1][j+1] = c[i][j]+1;
        else
          c[i+1][j+1] = c[i][j+1] > c[i+1][j] ? c[i][j+1] : c[i+1][j];
      }
    }
    if (c[xlen][ylen]==0) puts("0");
    else {
    for (i=18; c[xlen][ylen]>0; c[xlen][ylen]/=10) buf[i--]=(c[xlen][ylen]%10)+'0';
    puts(buf+i+1);
    }
  }
  return 0;
}