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.Arrays;
import java.math.*;
import java.lang.Long;
import java.util.Scanner;
public class ForTesting {

	public static void main(String[] args) {
		
		Scanner sc=new Scanner(System.in);
		
		int test=sc.nextInt();
	
		while(test>0)
		{
			int n=sc.nextInt();
			 int arr[]=new int[n];
			long sum=0; 
	          for(int i=0; i<n; i++)
	          { arr[i]=sc.nextInt(); sum+=arr[i]; }
	          if(sum==n) System.out.println(0);
	          else if(sum<n) System.out.println((sum%n)*((int)(Math.ceil((double)n/(double)sum))));
	          else if(sum>n && sum%n==0) {
	        	  
	        	  System.out.println(0);}
	          else 
	          {
	        	 long re=sum/n+1;
	        	  long res=sum%n;
	        	  long res1=sum-(re*res);
	        	long d=res1/(re-1);
	        	 long resi=d*res;
	        	  System.out.println(resi);
	          }
	          test--;
		}
		
  }
}