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: Wrong Answer)

#include<stdio.h>

int max(int a, int b) {

  if (a>b)
    return a;
  else
    return b;
}

int lcs(char *x, char *y){

  int i, j, xlen=0, ylen=0, c[1001][1001],maxl=0;


  for(i=0; i<=1000; i++)
    for(j=0; j<=1000; j++)
      c[i][j]=0;

  while (x[xlen])
    xlen++;


  while (y[ylen])
    ylen++;

  for (i=1; i<xlen; i++)
    c[i][0]=0;

  for (i=1; i<ylen; i++)
    c[0][i]=0;


  for(i=1; i<=xlen; i++)
    for(j=1; j<=ylen; j++){
      if (x[i] == y[j])
        c[i][j]=c[i-1][j-1]+1;
      else
        c[i][j]=max(c[i][j-1], c[i-1][j]);
        maxl=max(maxl, c[i][j]);
  }

return maxl;
}

int main() {

  char x[1001], y[1001];
  int i=0, j=0, k=0, p, c[1001][1001], ans[200];
  int xlen=0, ylen=0, maxl=0;
  scanf("%d", &p);

  for (k=0; k<p; k++){
    scanf("%s", x);
    scanf("%s", y);

    ans[k]=lcs(x, y);

  }
  for (i=0; i<p; i++)
    printf("%d\n", ans[i]);


  return 0;
}