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 fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(0);
#define endl "\n"
#define pb push_back
#define D(x) cout << "#" << x << endl;
#define LL long long
#define HI cout << endl \
                << "hi" << endl;
using namespace std;

// SOLUTION ..........
LL gcd(LL x, LL y) { return y == 0 ? x : gcd(y, x % y); }
void solv()
{
    LL n;
    cin >> n;
    LL a[n], sum = 0, i;
    for (i = 0; i < n; i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    LL q = sum % n;
    cout << q * (n - q) << endl;
}

// MAIN FUNCTION ..........
int main()
{
#ifndef ONLINE_JUDGE
    freopen("input1.txt", "r", stdin);
    freopen("output1.txt", "w", stdout);
#endif
    fast
        LL test = 1;
    cin >> test;
    while (test--)
    {
        solv();
    }
    return 0;
}
