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>
#define ll long long
#define db long double
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

using namespace std;

const int mod = 1000000007;

void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}

int mult(int a, int b) {
  return a * (ll)b % mod;
}

int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}

void solve() {
  ll n;
  cin >> n;

  ll sum = 0;
  for (int i = 0; i < n; ++i) {
    ll x;
    cin >> x;
    sum += x;
  }

  ll cnt = sum % n;
  cout << cnt * (n - cnt) << "\n";
}

int main(){
#ifdef LOCAL
	freopen("B_input.txt", "r", stdin);
	//freopen("B_output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(0);
	cin.tie(0);

  int t;
  cin >> t;

  for (int i = 0; i < t; ++i) {
    solve();
  }

}
