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>
#define N 1000

int lcs();
char X[N],Y[N];

int main(){
    int n,i;
    
    scanf("%d",&n);

    for(i=0;i<n;i++){
        scanf("%s%s",&X[1],&Y[1]);
        X[0]=' ';
        Y[0]=' ';
        printf("%d\n",lcs(X,Y));
    }

    return 0;
}

int lcs(){
    int c[N+1][N+1];
    int m=strlen(X)-1;
    int n=strlen(Y)-1;
    int i,j,max=0;

    for(i=1;i<=m;i++) {
        c[i][0]=0;
    }
    for(i=1;i<=n;i++){
        c[0][i]=0;
    }

    for(i=1;i<=m;i++){
        for(j=1;j<=n;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(max<c[i][j]) max=c[i][j];
        }
    }

    return max;
}