## 题目描述

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 aa where $a_i$ represents the number of traffic cars in the $i$ -th sub-track. You define the inconvenience of the track as $\sum\limits_{i=1}^{n} \sum\limits_{j=i+1}^{n} \lvert a_i-a_j\rvert$ , 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.

## 输入格式

The first line of input contains a single integer $t$ $( 1\leq t\leq 10\,000)$ — the number of test cases.

The first line of each test case contains a single integer $n$ $( 1\leq n\leq 2\cdot 10^5)$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ $( 0\leq a_i\leq 10^9)$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$ .

## 输出格式

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.

## 输入输出样例

### 输入 #1

3
3
1 2 3
4
0 1 1 0
10
8 3 6 11 5 2 1 7 10 4

### 输出 #1

0
4
21

## 说明/提示

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.

## 代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
int t,n,a;
int sum,op;
signed main(){
scanf("%lld",&t);
while(t--){
scanf("%lld",&n);
sum=0;
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
sum+=a[i];
}
op=sum%n;
printf("%lld\n",(n-op)*op);
}
}