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();
        for(int i=0;i<q;i++){
            String s = sc.next();
            String t = sc.next();
            int DP[][] = new int[s.length()+1][t.length()+1];
            for(int j=0;j<s.length()+1;j++)Arrays.fill(DP[j],0);
            for(int j=1;j<=s.length();j++){
                for(int k=1;k<=t.length();k++){
                    if(s.charAt(j-1)==t.charAt(k-1)) DP[j][k] = Math.max(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[s.length()][t.length()]);
        }
    }
}