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 <stdlib.h>
#include <string.h>


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


int main()
{
	int q;
	char x[1010], y[1010];
	int **table, *s_table;
	int x_len, y_len;
	int i, j, k;


	scanf("%d", &q);

	s_table = (int *)malloc(sizeof(int) * 1001 * 1001);
	table = (int **)malloc(sizeof(int *) * 1001);
	for(i = 0; i < 1001; i++)
		table[i] = s_table + (i * 1001);

	for(i = 0; i < q; i++){
		scanf("%s %s", x, y);
		x_len = strlen(x);
		y_len = strlen(y);

		// 初期化
		for(j = 0; j <= x_len; j++)
			table[0][j] = 0;
		for(j = 0; j <= y_len; j++)
			table[j][0] = 0;

		// LCSを求める
		for(j = 1; j <= x_len; j++){
			for(k = 1; k <= y_len; k++){
				if(x[j - 1] == y[k - 1]){
					table[k][j] = table[k - 1][j - 1] + 1;
				}else
					table[k][j] = max(table[k - 1][j], table[k][j - 1]);
			}
		}
		printf("%d\n", table[y_len][x_len]);
	}

	return 0;
}