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 <string>
#include <map>
#include <list>
#include <iterator>
#include <cmath>
#include <tuple>
#include <vector>
#include <algorithm>
#include <chrono>
#include <set>
#include <functional>
#include <utility>
#include <bitset>
#include <climits>
#include <numeric>

#define ll long long
#define ull unsigned long long int    
#define endl '\n'
#define PB push_back
#define vi vector<int>
#define vl vector<ll>
#define vvi vector<vi>
#define vvl vector<vl>

double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
ll M = 10e8 + 7;

using namespace std;


void solve()
{
	ll n; cin >> n;
	ll a, sum = 0;

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

	ll d1 = sum / n;
	ll d2 = sum / n;
	if (sum%n != 0)d2++;
	ll a1, a2;
	if (d1 == d2) 
	{
		cout << 0 << endl;
		return;
	}
	else
	{
		a1 = sum - n * d1;
		a2 = n * d2 - sum;
		if (a1 > a2) a1 = a2;
		
	}
	cout << (n - a1)*a1 << endl;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	ll t; cin >> t; while (t--)
	solve();

}