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 MAX 2000
#define max(a, b) ((a > b) ? a : b)

int w_long[MAX][MAX];//0に初期化するためにグローバルに置いてるだけ

int main(){
    int i, j;
    int n, wc1, wc2;
    char w1[MAX], w2[MAX];

    scanf("%d", &n);
    while(n--){
        scanf("%s%s", w1, w2);
        wc1 = strlen(w1);
        wc2 = strlen(w2);
        for(i = 0 ; i < wc1 ; i++){
            for(j = 0 ; j < wc2  ; j++){
                if(w1[i] == w2[j]){
                    w_long[i + 1][j + 1] = w_long[i][j] + 1;
                }else{
                    w_long[i + 1][j + 1] = max(w_long[i][j + 1], w_long[i + 1][j]);
                }
            }
        }
        printf("%d\n",w_long[wc1][wc2]);
    }
    return 0;
}