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.io.*;
public class Main {
	static public void main(String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(br.readLine());
		for (int i=0;i<n;i++){
			String sl=br.readLine();
			String ss=br.readLine();
			if (ss.length()>sl.length()){
				String tmp=sl;
				sl=ss;
				ss=tmp;
			}
			char[] cl=sl.toCharArray();
			char[] cs=ss.toCharArray();
			int[] length=new int[cs.length];
			int[] position=new int[cs.length];
			for (int j=0;j<cs.length;j++){
				position[j]=myfind(cl,cs[j],0);
				if (position[j]!=-1) length[j]=1;else length[j]=0;
				for (int k=0;k<j;k++){
					int tmpp=myfind(cl,cs[j],position[k]+1);
					if (tmpp!=-1){
						if (length[k]+1>length[j]){
							length[j]=length[k]+1;
							position[j]=tmpp;
						}else if(length[k]+1==length[j]&&tmpp<position[j]) position[j]=tmpp;
					}
				}
			}
			int max=0;
			for (int j=0;j<cs.length;j++) if (length[j]>max) max=length[j];
			System.out.println(max);
		}
	}
	public static int myfind(char[] c,char t,int p){
		for (;p<c.length;p++){
			if (c[p]==t) return p;
		}
		return -1;
	}
}