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;
int t,n; 
long long a; 
long long jumb = 0; 

long long rC2 (long long r) {
  return ((r)*(r-1))/2; 
}

int main () {
  scanf("%d", &t); 
  for(int q=0;q<t; q++){
    jumb = 0;
    scanf("%d", &n); 

    for(int i=0;i<n;i++){
      scanf("%lld", &a); 
      jumb += a; 
    } 
    long long hasilin0 = rC2(jumb%n) + rC2(n - jumb%n); 
    long long ans = rC2(n) - hasilin0; 
    
   // if(n==1) ans = a; 
    cout << ans << endl; 
  }
}