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<ctype.h>

int main(){
  
  int q, maxl, xi, yi, i, j, ii;
  char  X[1002], Y[1002];
  long int C[1001][1001], Z[150];
  
  scanf("%d", &q);
  
  for( ii = 0; ii < q; ii++){
    scanf("%s", X);
    scanf("%s", Y);
    /*????????°*/
    xi = 1;
    while(X[xi]){
      xi++;
    }
    yi = 1;
    while(Y[yi]){
      yi++;
    }
    
    C[0][0] = 0;
    for(i = 1; i <= xi; i++) {
    C[i][0] = 0;
    X[xi - i + 1] = X[xi - i];
    }
    for(i = 1; i <= yi; i++) {
    C[0][i] = 0;
    Y[yi - i + 1] = Y[yi - i];
    }
    X[0] = ' ';
    Y[0] = ' ';

    /* c[][]*/
    maxl = 0;
    for( i = 1; i <= xi; i++){
      for( j = 1; j <= yi; j++){
        if ( X[i] == Y[j] ){
          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];
          }
        }
        
        if( maxl < C[i][j]){
          maxl = C[i][j];
        }
      }
    }
    
    Z[ii] = maxl;
  }
  
  for(ii = 0; ii < q; ii++){
    printf("%d\n",Z[ii]);
  }
  
  return 0;
}