C. Raspberries

time limit per test

2 seconds

memory limit per test

256 megabytes

You are given an array of integers a1,a2,…,ana1,a2,…,an and a number kk (2≤k≤52≤k≤5). In one operation, you can do the following:

• Choose an index 1≤i≤n1≤i≤n,
• Set ai=ai+1ai=ai+1.

Find the minimum number of operations needed to make the product of all the numbers in the array a1⋅a2⋅…⋅ana1⋅a2⋅…⋅an divisible by kk.

Input

Each test consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains two integers nn and kk (2≤n≤1052≤n≤105, 2≤k≤52≤k≤5) — the size of the array aa and the number kk.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤101≤ai≤10).

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, output the minimum number of operations needed to make the product of all the numbers in the array divisible by kk.

Example

Input

Copy

15

2 5

7 3

3 3

7 4 1

5 2

9 7 7 3 9

5 5

5 4 1 2 3

7 4

9 5 1 5 9 5 1

3 4

6 3 6

3 4

6 1 5

3 4

1 5 9

4 4

1 4 1 1

3 4

3 5 3

4 5

8 9 9 3

2 5

1 6

2 5

10 10

4 5

1 6 1 1

2 5

7 7

Output

Copy

2
2
1
0
2
0
1
2
0
1
1
4
0
4
3

Note

In the first test case, we need to choose the index i=2i=2 twice. After that, the array will be a=[7,5]a=[7,5]. The product of all the numbers in the array is 3535.

In the fourth test case, the product of the numbers in the array is 120120, which is already divisible by 55, so no operations are needed.

In the eighth test case, we can perform two operations by choosing i=2i=2 and i=3i=3 in any order. After that, the array will be a=[1,6,10]a=[1,6,10]. The product of the numbers in the array is 6060.

解题说明：此题是一道数学题，由于K范围很小，可以分类讨论。当k=4的时候存在给两个数都加1的情况，其他情况下的k都只可以给一个数一直加1直到这个数可以整除k。

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int INF = 0x3f3f3f3f;
int main()
{
	int T; 
	cin >> T;
	while (T--)
	{
		int n, k; 
		cin >> n >> k;
		int a[100004];
		int ans = INF;
		int cal = 1;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			cal *= a[i];
			if (a[i] % k == 0) 
			{
				ans = 0;
			}
			int t = a[i] / k + 1;
			ans = min(ans, t * k - a[i]);
		}
		if (k == 4)
		{
			int cnt1 = 0, cnt2 = 0;
			for (int i = 1; i <= n; i++)
			{
				switch (a[i] % 4)
				{
				case 1:
					cnt1++;
					break;
				case 2:
					cnt2++;
					break;
				case 3:
					break;
				default:
					ans = 0;
					break;
				}
			}
			if (cnt1 >= 2)
			{
				ans = min((int)2, ans);
			}
			if (cnt2 >= 2)
			{
				ans = 0;
			}
			if (cnt1 >= 1 && cnt2 >= 1)
			{
				ans = min((int)1, ans);
			}
		}
		cout << ans << endl;
	}
	return 0;
}