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 1010

int memo[MAX][MAX];

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

void fill(int p, int q, char X[], char Y[]) {
	if (X[p - 1] == Y[q - 1]) {
		memo[p][q] = memo[p - 1][q - 1] + 1;
	}
	else {
		memo[p][q] = max(memo[p - 1][q], memo[p][q - 1]);
	}
}

int main() {
	int q;
	int x_len, y_len;
	char X[MAX], Y[MAX];

	for (int i = 0; i < MAX; i++) {
		memo[i][0] = 0;
		memo[0][i] = 0;
	}

	scanf("%d", &q);

	for (int i = 0; i < q; i++) {
		scanf("%s%s", X, Y);
		x_len = strlen(X); y_len = strlen(Y);
		for (int p = 1; p <= x_len; p++) {
			for (int q = 1; q <= y_len; q++) {
				fill(p, q, X, Y);
			}
		}
		printf("%d\n", memo[x_len][y_len]);


	}
	return 0;
}