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 C[1001][1001];

int main(){
  int n,i,j,k,l;
  char str1[1001],str2[1001];

   scanf("%d",&n);

   for(i = 0;i < n;i++){
     scanf("%s%s",str1,str2);

     int X,Y;
     X = strlen(str1);
     Y = strlen(str2);

     for(j = 0;j < X;j++){
       C[0][j] = 0;
     }
     for(j = 0;j < Y;j++){
       C[j][0] = 0;
     }

     for(k = 1;k <= Y;k++){
       for(l = 1;l <= X;l++){
         if(str1[l-1] == str2[k-1]) C[k][l] = C[k-1][l-1] + 1;
         else if(C[k][l-1] <= C[k-1][l]) C[k][l] = C[k-1][l];
         else C[k][l] = C[k][l-1];
       }
     }
     printf("%d\n",C[Y][X]);
   }
  return 0;
}