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 <stdlib.h>
#include <string.h>
#include <math.h>



int lcs[1001][1001];

char X[1001];
char Y[1001];


void makeLCS(int len_i, int len_j){
    
    int i, j;
    for (i = 0; i <= len_i; i++ ) {
        lcs[i][0] = 0;
    }
    
    for (j = 0; j <= len_j; j++ ) {
        lcs[0][j] = 0;
    }
    
    for (i = 1; i <= len_i; i++) {
        for(j = 1; j <= len_j; j++){
            if (X[i-1] == Y[j-1]) {
                lcs[i][j] = lcs[i-1][j-1] + 1;
            }
            else{
                if (lcs[i][j-1] > lcs[i-1][j]) {
                    lcs[i][j] = lcs[i][j-1];
                }
                else{
                    lcs[i][j] = lcs[i-1][j];
                }
            }
        }
    }
}


int main( ){
    
    int i, n;
    
    
    scanf("%d", &n);
    
    for (i=0; i<n; i++) {
        scanf("%s", X);
        scanf("%s",Y);
        makeLCS(strlen(X), strlen(Y));
        printf( "%d\n", lcs[strlen(X)][strlen(Y)] );
    }
    
    return 0;
}