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.lang.String;
 import java.util.Arrays;
 import java.lang.Math;
 
public class Main {
 
  
       
    public static void main(String[] args) {
 
        Scanner in = new Scanner(System.in);
        int i,j,k,l,m,t=0;
        t=in.nextInt();
        
        for(i=0;i<t;i++)
        {
             int n=in.nextInt();
             long o,p;
             long a[]=new long[n];
             long s1=0;
             for(j=0;j<n;j++)
             {a[j]=in.nextLong();
                 s1=s1+a[j];
                }
                o=s1%n;
                p=o*(n-o);
                System.out.println(p);
             
        }
        
}
       
}