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>

char X[1010], Y[1010];
int dp[1010][1010];

int LCS(){
    int maxl=0;
    int m, n, i, j;

    scanf("%s", &X);
    scanf("%s", &Y);
    m = strlen(X);
    n = strlen(Y);

    for(i=0; i<=m; i++){
        for(j=0; j<=n; j++){
            dp[i][j] = 0;
        }
    }
    for(i=0; i<m; i++){
        for(j=0; j<n; j++){
            if(X[i]==Y[j]){
                dp[i+1][j+1] = dp[i][j] +1;
            }else if(dp[i][j+1] < dp[i+1][j]){
                dp[i+1][j+1] = dp[i+1][j];
            }else{
                dp[i+1][j+1] = dp[i][j+1];
            }

            if(maxl < dp[i+1][j+1]){
                maxl = dp[i+1][j+1];
            }
        }
    }

    return maxl;
}

int main(){
    int q, i, j, k;
    int m, n;
    int ans;
    scanf("%d", &q);
    for(i=0; i<q; i++){
        ans = LCS();
        printf("%d\n", ans);
    }
    return 0;
}