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>          // printf(), scanf()
#include <string.h>         // strlen()

#define MAX_L 1000
#define max(a, b) ((a) > (b) ? (a) : (b))

int
main(int argc, char** argv)
{
	char s[MAX_L + 1], t[MAX_L + 1];
	int dp[MAX_L + 1][MAX_L + 1];
	int n;
	int i, j, k;

	scanf("%d", &n);
	for (k = 0; k < n; ++k)
	{
		scanf("%s", s);
		scanf("%s", t);
		int u = strlen(s);
		int v = strlen(t);

		for (i = 0; i <= u; ++i)
			dp[i][0] = 0;

		for (j = 0; j <= v; ++j)
			dp[0][j] = 0;

		for (i = 0; i < u; ++i)
		{
			for (j = 0; j < v; ++j)
			{
				if (s[i] == t[j])
					dp[i + 1][j + 1] = dp[i][j] + 1;
				else
					dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
			}
		}

		printf("%d\n", dp[u][v]);
	}

	return 0;
}