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<stdio.h>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<vector>
#include <unordered_map>
#include<climits>
#include<set>
#include <iomanip>
#include<numeric>
#include<queue>
#include<stack>
#include <map>
#include<bitset>
using namespace std;
#define ll long long
#define endl "\n"
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,x;
        cin>>n;
        ll sum=0;
        for(ll i=0;i<n;i++)
        {
            cin>>x;
            sum+=x;
        }
        cout<<(sum%n)*(n-(sum%n))<<endl;
    }
    return 0;
}