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++14 (judge verdict: Accepted)

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


int lcs(string a, string b) {
	int dp[1009][1009];

	for (int i = 0; i <= a.size(); i++) {
		for (int j = 0; j <= b.size(); j++) {
			if (i == 0 || j == 0) dp[i][j] = 0;
			else if (a[i - 1] == b[j - 1])
				dp[i][j] = max(max(dp[i - 1][j - 1] + 1, dp[i - 1][j]), dp[i][j - 1]);
			else
				dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
		}
	}
	return dp[a.size()][b.size()];
}

int main() {
	int q;
	cin >> q;
	getchar();
	while (q--) {
		string a, b;
		getline(cin, a);
		getline(cin, b);
		/*cout << "=======" << a << "=======" << endl;
		cout << "=======" << b << "=======" << endl;*/
		cout << lcs(a, b) << endl;
	}
	return 0;
}