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.*;
public class Main{
    public static void main(String[] args){
	Scanner sc=new Scanner(System.in);
	int q=sc.nextInt();
	int[] ans=new int[q];
	for(int n=0;n<q;n++){
	    String X=sc.next();
	    String Y=sc.next();
	    int[][] A=new int[X.length()+1][Y.length()+1];
	    for(int i=1;i<X.length()+1;i++){
		for(int j=1;j<Y.length()+1;j++){
		    if((X.substring(i-1,i)).equals(Y.substring(j-1,j))){
			A[i][j]=A[i-1][j-1]+1;
		    }else{
			if(A[i-1][j]<A[i][j-1]){
			    A[i][j]=A[i][j-1];
			}else{
			    A[i][j]=A[i-1][j];
			}
		    }
			
		}
	    }
	    ans[n]=A[X.length()][Y.length()];
	}
	for(int num:ans){
	    System.out.println(num);
	}
    }
}