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 {
	
	static int dp(String x, String y) {
		int[][] dp=new int[1010][1010];
		if(x.charAt(0)==y.charAt(0)) {
			dp[0][0]=1;
		}
		else if(x.charAt(0)!=y.charAt(0)) {
			dp[0][0]=0;
		}
		for(int i=1; i<x.length(); i++) {
			if(x.charAt(i)==y.charAt(0)) {
				dp[i][0]=1;
			}
			else {
				dp[i][0]=dp[i-1][0];
			}
		}
		for(int i=1; i<y.length(); i++) {
			if(y.charAt(i)==x.charAt(0)) {
				dp[0][i]=1;
			}
			else {
				dp[0][i]=dp[0][i-1];
			}
		}
		
		for(int i=1; i<x.length(); i++) {
			for(int j=1; j<y.length(); j++) {
				if(x.charAt(i)==y.charAt(j)) {
					dp[i][j]=Math.max(Math.max(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1])+1;
				}
				else {
					dp[i][j]=Math.max(Math.max(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]);
				}
			}
		}
		return dp[x.length()-1][y.length()-1];
	}
	
	public static void main(String[] args) {
		try(Scanner sc = new Scanner(System.in)){
			int q=sc.nextInt();
			String x, y;
			for(int k=0; k<q; k++) {
				x=sc.next();
				y=sc.next();
				System.out.println(dp(x, y));
			}
			

		}
	}
}