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 JAVA (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;


public class Main {

	private static int[][] cache;
	private static char[] x;
	private static char[] y;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in); 

		int set = sc.nextInt();

		for (int k = 0; k < set; k++) {
			x = sc.next().toCharArray();
			y = sc.next().toCharArray();
			int m = x.length;
			int n = y.length;
			
			// 再帰的トップダウン型、履歴管理
			cache = new int[m + 1][n + 1];
			for (int i = 1; i <= m; i++) {
				Arrays.fill(cache[i], Integer.MIN_VALUE);
				cache[i][0] = 0;
			}
			Arrays.fill(cache[0], 0);
			
			System.out.println(lcs(m, n));
		}
		
         sc.close();
	}

	private static int lcs(int m, int n) {
		if (cache[m][n] != Integer.MIN_VALUE) {
			return cache[m][n];
		}
		
		int result;
		if (x[m - 1] == y[n - 1]) {
			result = lcs(m - 1, n - 1) + 1;
		} else {
			result = Math.max(lcs(m - 1, n), lcs(m, n - 1));
		}
		
		cache[m][n] = result;
		return result;
	}
}