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.*;
import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int rslt[] = new int[n]; 
        for(int i=0;i<n;i++){
        	String A = sc.next();
			String B = sc.next();
			int[][] DP = new int[A.length() + 1][B.length() + 1];
			for(int j=0;j<A.length()+1;j++){//DP初期化
			    DP[j][0] = 0;
			}
	    	for(int j=0;j<B.length()+1;j++){
			    DP[0][j] = 0;
			}
			for(int j = 1; j <= A.length(); j++) {
				for(int k = 1; k <= B.length(); k++) {
					if(A.charAt(j - 1) == B.charAt(k - 1)) {
						DP[j][k] = DP[j - 1][k - 1] + 1;
					}else {
						DP[j][k] = Math.max(DP[j - 1][k], DP[j][k - 1]);
					}
				}
			}
			System.out.println(DP[A.length()][B.length()]);
        }
    }
}