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>
#include<stdlib.h>
char data[1010],data2[1010];
int alg(void);
int main(){
    int i,n,A[1010];
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%s %s",data,data2);
        A[i]=alg();
    }
    for(i=0;i<n;i++){
        printf("%d\n",A[i]);
    }
    return 0;
}
int alg(){
    int i,j,x,y,a[1010][1010];
    x=strlen(data);
    y=strlen(data2);
    for(i=0;i<=x;i++){
        a[i][0]=0;
    }
    for(j=0;j<=y;j++){
        a[0][j]=0;
    }
    for(i=1;i<=x;i++){
        for(j=1;j<=y;j++){
            if(data2[j-1]==data[i-1])a[i][j]=a[i-1][j-1]+1;
            else if(a[i-1][j]>=a[i][j-1])a[i][j]=a[i-1][j];
            else a[i][j]=a[i][j-1];
        }
    }
    return a[x][y];
}