Polycarp has n friends, the i-th of his friends has a_i candies. Polycarp's friends do not like when they have different numbers of candies. In other words they want all a_i to be the same. To solve this, Polycarp performs the following set of actions exactly once: 

  * Polycarp chooses k (0 ≤ k ≤ n) arbitrary friends (let's say he chooses friends with indices i_1, i_2, …, i_k); 
  * Polycarp distributes their a_{i_1} + a_{i_2} + … + a_{i_k} candies among all n friends. During distribution for each of a_{i_1} + a_{i_2} + … + a_{i_k} candies he chooses new owner. That can be any of n friends. Note, that any candy can be given to the person, who has owned that candy before the distribution process. 



Note that the number k is not fixed in advance and can be arbitrary. Your task is to find the minimum value of k.

For example, if n=4 and a=[4, 5, 2, 5], then Polycarp could make the following distribution of the candies: 

  * Polycarp chooses k=2 friends with indices i=[2, 4] and distributes a_2 + a_4 = 10 candies to make a=[4, 4, 4, 4] (two candies go to person 3). 



Note that in this example Polycarp cannot choose k=1 friend so that he can redistribute candies so that in the end all a_i are equal.

For the data n and a, determine the minimum value k. With this value k, Polycarp should be able to select k friends and redistribute their candies so that everyone will end up with the same number of candies.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^4).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case output: 

  * the minimum value of k, such that Polycarp can choose exactly k friends so that he can redistribute the candies in the desired way; 
  * "-1" if no such value k exists. 

Example

Input


5
4
4 5 2 5
2
0 4
5
10 8 5 1 4
1
10000
7
1 1 1 1 1 1 1


Output


2
1
-1
0
0

#include<bits/stdc++.h>

using namespace std;


#define ff              first
#define ss              second
#define int             long long
#define pb              push_back
#define mp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define all(a)      (a).begin(),(a).end()

#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
#define REP(i,n) for(i=0;i<n;i++)
#define FIO             ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

const long long   M = 1e9 + 7;

void c_p_c()
{
#ifndef  ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("o.txt", "w", stdout);
#endif
}

int mod(int x) {
	return (x % M + M) % M;
}
int  add(int a, int b)
{
	return mod(mod(a) + mod(b));
}
bool comp(pii a, pii b)
{

	return  a.ss < b.ss;
}
bool f(int x, int y)
{
	return x > y;
}
void solve()
{
	int n;
	cin >> n;
	int a[n];
	int i;
	int s = 0;
	for (i = 0; i < n; i++)
	{
		cin >> a[i];
		s += a[i];
	}
	if (s % n != 0)
	{
		cout << "-1\n";
		return;
	}
	int res = s / n;
	int c = 0;
	for (i = 0; i < n; i++)
	{
		if (a[i] == res)
			continue;

		else if (a[i] > res)
		{
			c++;
		}
	}
	cout << c << endl;


}


int32_t main()
{

	c_p_c();

	FIO;

	int tt = 1;
	w(t)
	solve();



}






