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.
/// #514
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int > pii;
typedef long long int lld;
#define pi acos(-1)
#define fr(i,m,n) for(i=m;i<n;i++)
#define fu(i,m,n) for(i=m;i>=n;i--)
#define vec vector<lld>
#define pb push_back
#define pp pop_back()
#define ft first
#define sd second
#define all(v) v.begin(),v.end()
#define mom(ara) memset(ara,0,sizeof(ara));
#define m1m(ara) memset(ara,-1,sizeof(ara));
#define endl "\n"
#define eps 1.19209e-07
int main()
{ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);
    lld i,j,k,a,b,c,x,y,z,n,m,ans,t;
    cin>>t;
    while(t--)
    {
    	cin>>n;lld sum=0;
    	fr(i,0,n)
    	{
    		cin>>x;
    		sum+=x;
    	}
    	sum%=n;
    	cout<<(n-sum)*sum<<endl;
    }
}