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.io.*; 
import java.util.Scanner;
import java.util.*;
import java.lang.*;
import java.io.IOException;
public class Codechef
{ 
    public static void main(String[] args) throws IOException
    {
        try 
        {
            System.setIn(new FileInputStream("input.txt"));
            System.setOut(new PrintStream(new FileOutputStream("output.txt")));
        } 
        catch (Exception e) 
        {
            System.err.println("Error");
        }
      
         Scanner sc= new Scanner(System.in);
         int t = sc.nextInt();
         while(t--!=0)
         {
            int n = sc.nextInt();
            int[] a = new int[n];
            long sum = 0;
            for(int i=0;i<n;i++)
            {
                a[i] = sc.nextInt();
                sum = sum + a[i];
            }
          //  System.out.println(sum);

            if(sum%n==0)
            {
                System.out.println(0);
            }
            else
            {
                long temp = (long)Math.ceil(sum/n);
                long res  = sum - temp*n;
                System.out.println((n-res)*res);
            }
            
         }

    }

    

    
}
