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 <vector>
#include <iostream>
#include <string>
#include <algorithm>
#include <iterator>
using namespace std;
int main(){
	int n;
	cin >> n;
	while(n--){
		string a,b;
		cin >> a >> b;
		int asize = a.size(),bsize = b.size();
		vector<vector<int> > dp(asize+1,vector<int>(bsize+1,0));
		for(int i=1;i<asize+1;i++){
			for(int j=1;j<bsize+1;j++){
				if(a[i-1] == b[j-1]){
					dp[i][j] = dp[i-1][j-1] + 1;
				}else{
					dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
				}
			}
		}
		/*				for(int i = 0;i<asize+1;i++){
		copy(dp[i].cbegin(),dp[i].cend(),ostream_iterator<int>(cout, " ") );
		cout << endl;
		}*/
				
		cout << dp[asize][bsize] << endl;
	}
}