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 MAX 1000

int max(int a, int b)
{
	if(a>b) return a;
	else return b;
}

int main(void)
{
	int i, j, q, Xlen, Ylen, l[MAX+1][MAX+1];
	char X[MAX+1], Y[MAX+1];
	
	scanf("%d",&q);
	while (q--) {
		scanf("%s",X);
		scanf("%s",Y);
		Xlen = strlen(X);
		Ylen = strlen(Y);
		for (i=0; i<=Xlen; i++) {
			for (j=0; j<=Ylen; j++) {
				if (i==0 || j==0) {
					l[i][j] = 0;
				} else if (X[i-1] == Y[j-1]) {
					l[i][j] = l[i-1][j-1] + 1;
				} else {
					l[i][j] = max(l[i-1][j], l[i][j-1]);
				}
			}
		}
		printf("%d\n", l[Xlen][Ylen]);
	}
	
	return 0;
}