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 N 1000

int lcs (char X[], char Y[]) {
	int i, j;
	int c[N + 1][N + 1];
	int m = strlen(X);
	int n = strlen(Y);
	int maxl = 0;


	for ( i = 1; i <= m; i++) c[i][0] = 0;
	for ( j = 1; 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 {
				c[i][j] = c[i - 1][j] > c[i][j - 1] ? c[i - 1][j] : c[i][j - 1] ;
			}
			maxl = maxl > c[i][j] ? maxl : c[i][j];
		}
	}

	return maxl;
}

int main () {
	char s1[N], s2[N];
	int n, i, result[N];

	scanf("%d",&n);
	for ( i = 0; i < n; i++ ) {
		scanf(" %s %s", s1, s2);
		result[i] = lcs(s1,s2);
		printf("%d\n",lcs(s1,s2));
	}

	return 0;
}