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++11 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

template<class T> inline void chmax(T &a, const T &b) { if(a < b) a = b; }

int main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int q;
	cin >> q;

	while(q--) {
		string a, b;
		cin >> a >> b;

		const int n = a.size();
		const int m = b.size();

		vector<vector<int>> dp(n + 2, vector<int>(m + 2, 0));
		for(int i = 0; i <= n; ++i) {
			for(int j = 0; j <= m; ++j) {
				if(a[i] == b[j]) {
					chmax(dp[i + 1][j + 1], dp[i][j] + 1);
				}
				else {
					chmax(dp[i + 1][j + 1], dp[i][j]);
					chmax(dp[i + 1][j], dp[i][j]);
					chmax(dp[i][j + 1], dp[i][j]);
				}
			}
		}

		cout << dp[n][m] << endl;
	}

	return EXIT_SUCCESS;
}