Highway 201 is the most busy street in Rockport. Traffic cars cause a lot of hindrances to races, especially when there are a lot of them. The track which passes through this highway can be divided into n sub-tracks. You are given an array a where a_i represents the number of traffic cars in the i-th sub-track. You define the inconvenience of the track as ∑_{i=1}^{n} ∑_{j=i+1}^{n} \lvert a_i-a_j\rvert, where |x| is the absolute value of x. 

You can perform the following operation any (possibly zero) number of times: choose a traffic car and move it from its current sub-track to any other sub-track.

Find the minimum inconvenience you can achieve.

Input

The first line of input contains a single integer t (1≤ t≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 2⋅ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (0≤ a_i≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Output

For each test case, print a single line containing a single integer: the minimum inconvenience you can achieve by applying the given operation any (possibly zero) number of times.

Example

Input


3
3
1 2 3
4
0 1 1 0
10
8 3 6 11 5 2 1 7 10 4


Output


0
4
21

Note

For the first test case, you can move a car from the 3-rd sub-track to the 1-st sub-track to obtain 0 inconvenience.

For the second test case, moving any car won't decrease the inconvenience of the track.
#include<bits/stdc++.h>
using namespace std;

long long int mod(long long int a)
{
	if(a<0)
		return -a;
	return a;
}

int main()
{
	long long int t;
	cin>>t;
	while(t--)
	{
		long long int n, x=0, sum=0;
		cin>>n;
		vector<long long int> a;
		for(long long int i=0;i<n;i++)
		{
			long long int el;
			cin>>el;
			a.push_back(el);
			sum+=el;
		}
		long long int d, r;
//		d=sum/n;
		r=sum%n;
//		for(long long int i=0;i<n;i++)
//			a[i]=d;
//		for(long long int i=0;i<r;i++)
//			a[i]++;
		d=n-r;
		x=d*r;	
//		for(long long int i=0;i<n;i++)
//		{
//			for(long long int j=i;j<n;j++)
//			{
//				x+=mod(a[i]-a[j]);
//			}
//		}
		cout<<x<<endl;
	}
}
