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 MAXLEN 1001
#define MAX(a, b) ((a) < (b) ? (b) : (a))

int
lcs(char *xs, int _xl, char *ys, int _yl)
{
	int i, j;
	int xl = _xl + 1, yl = _yl + 1;
	static int *buf = NULL;
	static int buflen = 0;

	if (buflen < xl * yl) {
		if (buf != NULL)
			free(buf);
		buf = (int *)malloc(sizeof(int) * xl * yl);
		if (buf == NULL)
			return -1;
		buflen = xl * yl;
	}

#define BUF(i, j) buf[(j) * xl + (i)]
	for (j = 0; j < yl; j++)
		BUF(0, j) = 0;

	for (i = 1; i < xl; i++) {
		BUF(i, 0) = 0;
		for (j = 1; j < yl; j++) {
			if (xs[i-1] == ys[j-1])
				BUF(i, j) = BUF(i-1, j-1) + 1;
			else
				BUF(i, j) = MAX(BUF(i, j-1), BUF(i-1, j));
		}
	}
#ifdef DEBUG
	for (j = 0; j < yl; printf("\n"), j++)
		for (i = 0; i < xl; i++)
			printf("%d ", BUF(i, j));
#endif

	return BUF(xl-1, yl-1);
#undef BUF
}

int
main(void)
{
	int i, n, xl, yl, v;
	char xs[MAXLEN];
	char ys[MAXLEN];

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%s%s", xs, ys);
		xl = strlen(xs);
		yl = strlen(ys);
		if (xl < yl)
			v = lcs(xs, xl, ys, yl);
		else
			v = lcs(ys, yl, xs, xl);
		if (v < 0)
			return 1;
		printf("%d\n", v);
	}


	return 0;
}