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 mp make_pair
#define pb push_back
#define SZ(x) (int)(x.size())
#define all(x) x.begin(),x.end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int maxn = 2e6 + 6;
const int  inf = 0x3f3f3f3f;
const int mod = 998244353;
const int N = 2e3 + 3;
ll qpow(ll x,ll y) { ll ans=1;x%=mod;  while(y) { if(y&1) ans=ans*x%mod; x=x*x%mod;  y>>=1; } return ans; }
//
int _;

int n;
ll a,b;

void so() {
    scanf("%d",&n);ll s=0;
    for(int i=1;i<=n;i++) {
        int d;
        scanf("%d",&d);
        s+=d;
    }
    printf("%lld\n",s%n*(n-s%n));

}
void init() {

}
int main() {
    init();
    for(scanf("%d",&_);_;_--) { 
        so();
    }
    return 0;
}