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
#define Q 150

int dp[N][N];

int match(char m, char n){
	if(m==n) return 1;
	else return 0;
}

int max(int a, int b, int c){
	int max = a;
	if(max<b) max = b;
	if(max<c) max = c;

	return max;
}

int main(void)
{
	int i, j, k, q, m, n, cnt;
	char X[N], Y[N];
	int Z[N];

	scanf("%d", &q);

	for(k=0; k<q; k++){
		scanf("%s", X); m = strlen(X);
		scanf("%s", Y); n = strlen(Y);

		// 初期化

		cnt = 0;

		for(i=0; i<m; i++){
			for(j=0; j<n; j++){
				dp[i][j] = 0;
			}
		}

		for(i=0; i<m; i++){
			if(cnt == 0 && match(X[i], Y[0])){
				cnt++;
			}
			dp[i][0] = cnt;
		}
		
		cnt = 0;

		for(j=0; j<n; j++){
			if(cnt == 0 && match(X[0], Y[j])){
				cnt++;
			}
			dp[0][j] = cnt;
		}

		//初期化終わり

		for(i=1; i<m; i++){
			for(j=1; j<n; j++){
				dp[i][j] = max(dp[i-1][j-1] + match(X[i], Y[j]), dp[i][j-1], dp[i-1][j]);
			}
		}

		Z[k] = dp[m-1][n-1];
	}

	for(i=0; i<q; i++){
		printf("%d\n", Z[i]);
	}

	return 0;
}