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>
#define N 1001

void Ics(char *X, char *Y)
{
   int c[N][N];
   int m, n, i, j;

   m = strlen(X);
   n = strlen(Y);

   for ( i = 0; i <= m; i++ ) {
      c[i][0] = 0;
   }
   for ( j = 1; j <= n; j++ ) {
      c[0][j] =0;
   }

   for ( i = 1; i <= m; i++ ) {
      for ( j = 1; j <=n; 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 ];
         }
      }
   }
   printf("%d\n", c[ m ][ n ]);
}
     
int main()
{
   int n, i;
   char X[N], Y[N];

   scanf("%d", &n);
   for ( i = 0; i < n; i++ ) {
      scanf("%s%s", X, Y);
      Ics(X, Y);
   }
   return 0;
}