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: WA: Presentation Error)

import java.util.Scanner;


public class Main{

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int cnt = Integer.parseInt(s.nextLine());
		System.out.println("");
		for(int i=0 ; i<cnt ; i++) {
			char[] a = s.next().toCharArray();
			char[] b = s.next().toCharArray();
			System.out.println(solve(a, b));
		}
	}

	public static int solve(char[] a, char[] b) {
		int[][] result = new int[a.length+1][b.length+1];
		for(int i=1 ; i<=b.length ; i++) {
			for(int j=1 ; j<=a.length ; j++) {
				if(a[j-1]==b[i-1]) {
					result[j][i] = result[j-1][i-1] + 1;
				}else {
					result[j][i] = Math.max(result[j][i-1], result[j-1][i]);
				}
			}
		}
		return result[a.length][b.length];
	}

}