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<iostream>
#include<bits/stdc++.h>
#include <cmath>

#define ll long long int
#define mp make_pair
#define pb push_back
#define MOD 1000000007
#define fi first
#define se second
#define inf INT_MAX

#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL);
#define all(v) v.begin(),v.end()

#define sumof(v) accumulate(all(v), 0LL, plus<ll>())
#define maxof(v) max_element(all(v))
#define minof(v) min_element(all(v))


#define Vl vector<ll>
#define Vlp vector<pair<ll,ll>>
#define Vll vector<vector<ll>>
#define Vi vector<int>
#define Vip vector<pair<int,int>>
#define Vii vector<vector<int>>

#define repU(i,x,y,d) for(int i=x;i<y;i+=d)
#define repD(i,x,y,d) for(int i=x;i>=y;i-=d)

using namespace std;

int n;

int main(){
    FASTIO
    int t=1;
    cin>>t;

    while(t--){
        ll val;
        cin>>n;
        ll sm=0;
        repU(i,0,n,1){cin>>val;sm+=val;}

        if(sm%n){
            cout<<(sm%n)*(n-(sm%n))<<'\n';
        }
        else cout<<0<<'\n';

    }
    return 0;
}
