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.Scanner;

public class Main {
	
	public static void main(String args[]) {
		
		Scanner input = new Scanner(System.in);
		int numberOfDatasets = input.nextInt();
		
		for(int i = 0; i < numberOfDatasets; i++) {
			
			String first = input.next();
			String second = input.next();
			
			System.out.println(compare(first, second));
			
		}
		
	}
	
	public static int compare(String first, String second) {
		
		String firstBlank = " " + first;
		String secondBlank = " " + second;
		
		int length1 = firstBlank.length();
		int length2 = secondBlank.length();
		
		
		int[][] LCS = new int[length1][length2];
		
		for(int i = 1; i < length1; i++) {
			
			for(int j = 1; j < length2; j++) {
				
				char one = firstBlank.charAt(i);
				char two = secondBlank.charAt(j);
				
				if(one == two) {
					
					LCS[i][j] = 1 + LCS[i - 1][j - 1];
					
				}else {
					
					LCS[i][j] = Math.max(LCS[i-1][j], LCS[i][j-1]);
					
				}
				
			}
			
		}
		
		
		return LCS[length1 - 1][length2 - 1];
		
	}

}