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 C++ (judge verdict: Accepted)

// LCSの練習

#include"bits/stdc++.h"
#include <iostream>
#include <vector>


using namespace std;


int main(){
	
	int q;

	cin >> q ;

	vector<int> ans;

	for(int i= 0; i < q; i++){
		
		string x,y;

		cin >> x;
		cin >> y;

		int len_x = (int)x.length();
		int len_y = (int)y.length();

		// int dp[len_x+1][len_y+1];

		vector<vector<int> > dp(len_x + 1, vector<int>(len_y + 1));

		for(int i = 1; i <= len_x; i++){
			for(int j = 1; j <= len_y; j++){
				if(x[i-1] == y[j-1]){
					dp[i][j] = dp[i-1][j-1]+1;
				}
				else{
					dp[i][j] = max(dp[i-1][j], dp[i][j-1]); 
				}
			}
		}

		ans.push_back(dp[len_x][len_y]);
		// cout << dp[len_x][len_y] << endl;

	}

	for(int i = 0; i < q; i++){
		cout << ans[i] << endl;
	}

	return 0;
}