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;

 
typedef long long ll; 
typedef long double lld;
typedef unsigned long long ull;

 
//ll n, m, q, k, l, r, x, y, z;
const ll template_array_size = 1e6 + 16614;
// ll a[template_array_size];
// ll b[template_array_size];
// ll c[template_array_size];
// string s, t;
 

void solve(int tc = 0) {
	
	ll n,x;
	cin>>n;
	ll sum=0;
	for(int i=0;i<n;++i)
	{
		cin>>x;
		sum+=x;

	}
	sum%=n;
	cout<<sum*(n-sum)<<endl;

	
}
 
int main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	// #ifndef ONLINE_JUDGE
	// freopen("input.txt","r",stdin);
	// freopen("output.txt","w",stdout);
	// #endif
	
		
	int tc = 1;
	cin >> tc; 	
	for (int t = 0; t < tc; t++) solve(t);

	
	return 0;

}