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.
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#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <math.h>
#include <string>
#include <cstring>
#include <cassert>
using namespace std;
 
const double PI = acos(-1);
const int N=1e5 + 10;
const int M=1000000007;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define oo 1000000001

void solve(){
    int n;
    cin>>n;
    vi a(n);
    ll sum=0;
    for (int i = 0; i < n; ++i)
    {
        cin>>a[i];
        sum+=a[i];
    }
    ll b=sum%n;
    printf("%lld\n",b*(n-b) );

}
 
int main(){
	int t=1;
	scanf("%d",&t);
	while(t--)
		solve();

	return 0;
}