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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		try(Scanner scan = new Scanner(System.in)) {
			int n = scan.nextInt();
			scan.nextLine();
			for(int i=0; i<n; i++) {
				String x = scan.nextLine();
				String y = scan.nextLine();
				int ans = solve(x, y);
				System.out.println(ans);
			}
		}
	}
	
	private static int solve(String x, String y) {
		int[][] dp = new int[x.length()+1][y.length()+1];
		for(int i=1; i<=x.length(); i++) {
			for(int j=1; j<=y.length(); j++) {
				boolean match = x.charAt(i-1) == y.charAt(j-1);
				dp[i][j] = Math.max(dp[i][j-1], (dp[i-1][j] + (match ? 1 : 0)));
				if(dp[i][j] > j) dp[i][j] = j;
			}
		}
		return dp[x.length()][y.length()];
	}
}