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>

void reset(int dp[][1000]){
    int i,j;
    for(i=0;i<1000;i++){
        for(j=0;j<1000;j++){
            dp[i][j]=0;
        }
    }
}

int main(void){
    int n,i,j,k,alen,blen;
    int dp[1001][1001];
    char a[1024],b[1024];
    scanf("%d",&n);
    for(k=0;k<n;k++){
        reset(dp);
        scanf("%s",a);
        scanf("%s",b);
        alen =strlen(a);blen =strlen(b);
        for(i=1;i<=alen;i++){
            for(j=1;j<=blen;j++){
                if(a[i-1]==b[j-1]){
                    dp[i][j]=dp[i-1][j-1]+1;
                }else{
                    dp[i][j]=dp[i-1][j];
                    if(dp[i][j]<dp[i][j-1]){
                        dp[i][j]=dp[i][j-1];
                    }
                }
            }
        }
        printf("%d\n",dp[alen][blen]);
    }
}