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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
		try(Scanner sc = new Scanner(System.in)) {
			int n = sc.nextInt();
			while(n-->0) {
				String s1 = sc.next();
				String s2 = sc.next();
				int len1 = s1.length();
				int len2 = s2.length();
				int[][] dp = new int[len1][len2];
				for(int i=0; i<len1; i++) {
					for(int j=0; j<len2; j++) {
						dp[i][j] = 0;
					}
				}
				for(int i=0; i<len1; i++) {
					dp[i][0] = (s1.charAt(i)==s2.charAt(0))? 1: 0;
				}
				for(int j=0; j<len2; j++) {
					dp[0][j] = (s2.charAt(j)==s1.charAt(0))? 1: 0;
				}
				for(int i=0; i<len1; i++) {
					for(int j=0; j<len2; j++) {
						for(int ni=0; ni<i; ni++) {
							for(int nj=0; nj<j; nj++) {
								if(s1.charAt(i) == s2.charAt(j))
									dp[i][j] = Math.max(dp[i][j], dp[ni][nj]+1);
							}
						}
					}
				}
				int ans = 0;
				for(int i=0; i<len1; i++) {
					for(int j=0; j<len2; j++) {
						ans = Math.max(ans, dp[i][j]);
					}
				}
				System.out.println(ans);
			}
		}
	}

}