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)

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int dp[1005][1005] = { 0 };
int main() {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		string x, y;
		cin >> x >> y;
		for (int j = 0; j < 1005; j++)
			for (int k = 0; k < 1005; k++) dp[j][k] = 0;
		for (int j = 1; j <= x.size(); j++) {
			for (int k = 1; k <= y.size(); k++) {
				dp[j][k] = max(dp[j][k - 1], dp[j - 1][k]);
				if (x[j - 1] == y[k - 1]) {
					dp[j][k] = max(dp[j][k], dp[j - 1][k - 1] + 1);
				}
			}
		}
		cout << dp[x.size()][y.size()] << endl;
	}
	return 0;
}