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: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>

using namespace std;

string s, t;

int main() {
	int n;
	cin >> n;
	while (n--) {
		//dp[i][j] := s???i??????????????§???t???j??????????????§?????£?????¨??????LCS
		vector<vector<int>> dp(2000, vector<int>(2000));
		cin >> s >> t;
		dp[0][0] = (s[0] == t[0]);
		for (int i = 1; i < s.size(); i++)dp[i][0] = (bool)(dp[0][0] + (s[i] == t[0]));
		for (int j = 1; j < t.size(); j++)dp[0][j] = (bool)(dp[0][0] + (s[0] == t[j]));
		
		for (int i = 1; i < s.size(); i++) {
			for (int j = 1; j < t.size(); j++) {
				if (s[i] == t[j])dp[i][j] = dp[i-1][j-1]+1;
				else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
			}
		}
		cout << dp[s.size() - 1][t.size() - 1] << endl;
	}
}