#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
	int n, m, q, A[1001][1001], i, j;
	string str1, str2;
	cin >> q;

	while (q--)
	{
		cin >> str1 >> str2;
		n = str1.size();
		m = str2.size();
		int maxv = 0;

		for (i = 1; i <= n; i++)A[i][0] = 0;
		for (i = 1; i <= m; i++)A[0][i] = 0;

		for(i=1;i<=n;i++)
			for (j = 1; j <= m; j++)
			{
				if (str1[i-1] == str2[j-1])
					A[i][j] = A[i - 1][j - 1] + 1;
				else A[i][j] = max(A[i - 1][j], A[i][j - 1]);
			}
		cout << A[n][m] << endl;
	}
	return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

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

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.

Example

Input

3
abcbdab
bdcaba
abc
abc
abc
bc


Output

4
3
2

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
	int n, m, q, A[1001][1001], i, j;
	string str1, str2;
	cin >> q;

	while (q--)
	{
		cin >> str1 >> str2;
		n = str1.size();
		m = str2.size();
		int maxv = 0;

		for (i = 1; i <= n; i++)A[i][0] = 0;
		for (i = 1; i <= m; i++)A[0][i] = 0;

		for(i=1;i<=n;i++)
			for (j = 1; j <= m; j++)
			{
				if (str1[i-1] == str2[j-1])
					A[i][j] = A[i - 1][j - 1] + 1;
				else A[i][j] = max(A[i - 1][j], A[i][j - 1]);
			}
		cout << A[n][m] << endl;
	}
	return 0;
}

```