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.
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>

int min ( int a, int b ) {
	if ( a <= b ) return a ;
	return b ;
}

int max ( int a, int b ) {
	if ( a >= b ) return a ;
	return b ;
}

int main () {
	int t ;
	scanf("%d", &t) ;
	long long int res[t] ;

	for ( int i=0 ; i<t ; ++i) {
		long long int n ;	scanf("%lld", &n) ;
		long long int ary[n] ; 
		for ( int j=0 ; j<n ; ++j ) scanf ("%lld", &ary[j]) ;

		long long int sum = 0 ;
	    for ( int j=0 ; j<n ; ++j ) {
	    	ary[j] = ary[j] % n ;
	    	sum += ary[j] ;
	    }

	    long long int remainder = sum % n ;
	    res[i] = remainder*( n-remainder ) ;
	}

	for ( int i=0 ; i<t ; ++i) {
		printf("%lld\n", res[i] );
	}
}