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: Time Limit Exceeded)

#include <stdio.h>

int main() {	
	char X[150][1000];
	char Y[150][1000];
	int figureXY[1001][1001] = { 0 };
	int longX, longY;
	int dataSetN;
	int max;
	int i, j, k;
	longX = 0;	longY = 0;
	max = 0;

	scanf("%d", &dataSetN);
	for (i = 0; i < dataSetN; i++) {
		scanf("%s", X[i], 2000);
		scanf("%s", Y[i], 2000);
	}

	for (i = 0; X[i] != '\n'; i++) {
		longX++;
	}
	for (i = 0; Y[i] != '\n'; i++) {
		longY++;
	}
	for (i = 0; i <= longX; i++) {
		figureXY[i][0] = 0;
	}
	for (i = 0; i <= longY; i++) {
		figureXY[0][i] = 0;
	}

	for (i = 0; i < dataSetN; i++) {
		max = 0;
		for (j = 1; j <= longX; j++) {
			for (k = 1; k <= longY; k++) {
				if (X[i][j - 1] == Y[i][k - 1]) {
					figureXY[j][k] = figureXY[j - 1][k - 1] + 1;
				}
				else if (figureXY[j - 1][k] >= figureXY[j][k - 1]) {
					figureXY[j][k] = figureXY[j - 1][k];
				}
				else {
					figureXY[j][k] = figureXY[j][k - 1];
				}

				max = figureXY[j][k] > max ? figureXY[j][k] : max;
			}
		}

		printf("%d\n", max);
	}

	return 0;
}