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;
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define int long long
#define ld long double
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rep(i,x,y) for(int i=x; i<y; i++)
#define fill(a,b) memset(a,b,sizeof(a))
#define vi vector<int>
#define vii vector<pair<int,int>>
#define setbits(x) __builtin_popcountll(x)
#define w(x)  int x; cin>>x; while(x--)
#define in(a) for(auto &x: a)cin>>x;
#define out(a) for(auto x: a){cout<<x<<' ';}cout<<'\n';
#define inf 1000000000
//#define mm 1000000007 998244353
void print(bool n){if(n)cout<<"YES";else cout<<"NO";cout<<'\n';}
signed main(){
	fastio
	w(t){
		int s=0;
		int n;
		cin>>n;
		int a[n];
		in(a);
		rep(i,0,n)s+=a[i];
		int x=(s/n)*n;
		int r;
		if(x==0)r=s;
		else r=s%x;
		cout<<(int)r*(n-r)<<'\n';
	}
	return 0;
}