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)

// AOJ ALDS1_10_C Longest Common Subsequence
// 2018.4.27 bal4u

#include <stdio.h>
#include <stdlib.h>

// Longest Common Subsequence
int c[1005][1005];
int LCS(char *x, int m, char *y, int n)
{
	int i, j;

	for (i = 0; i <= m; i++)  c[i][0] = 0;
	for (j = 0; j <= n; j++)  c[0][j] = 0;

	for (i = 1; i <= m; i++) for (j = 1; j <= n; j++) {
		if (x[i-1] == y[j-1]) c[i][j] = c[i-1][j-1] + 1;
		else if (c[i-1][j] >= c[i][j-1]) c[i][j] = c[i-1][j];
		else                             c[i][j] = c[i][j-1];
	}
	return c[m][n];
}

char x[1005], y[1005];

int main()
{
	int q, lx, ly;
	char *p;

	fgets(x, 10, stdin);
	q = atoi(x);
	while (q--) {
		fgets(p=x, 1005, stdin);
		while (*p > ' ') p++;
		*p = 0, lx = p-x;

		fgets(p=y, 1005, stdin);
		while (*p > ' ') p++;
		*p = 0, ly = p-y;

		printf("%d\n", LCS(x, lx, y, ly));
	}
	return 0;
}