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.

import java.util.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(
				new InputStreamReader(System.in));	 
		StringTokenizer st
		= new StringTokenizer(br.readLine());
		int i,n,t,p;
		long k,j,l,m;
		t = Integer.parseInt(st.nextToken());
		for(p=1;p<=t;p++)
		{   
			st = new StringTokenizer(br.readLine());
		    n = Integer.parseInt(st.nextToken());
		    long a[] = new long[n];
		    k = 0;
		    st = new StringTokenizer(br.readLine());
		    for(i=0;i<n;i++)
	        {
		    	a[i]=Long.parseLong(st.nextToken());
	            k+=a[i];
	        }
		    l = k%n;
		    if(l==0)
		    {
		    	System.out.println(0);
		     continue;
		    }
		    	       
		 System.out.println(((l)*(n-l)));
	}	
	}
}
