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 int          long long
#define pb           push_back
#define lld          long double
#define all(c)       c.begin(),c.end()
#define si(x)        (int)(x.size())
#define pii          pair<int,int>
#define ff           first
#define ss           second
#define FOR(x,a)     for(auto&x :a)
#define rep(i,a,b)   for(int i=a;i<b;i++)
#define repd(i,a,b)  for(int i=a;i>=b;i--)
#define mem(a,x)     memset(a,x,sizeof(a))
#define pnr(a)       FOR(i,a){cout<<i<<" ";}cout<<"\n";

#define endl '\n'
const long long inf = 1e18;
const int32_t mod = 1000000007;

#ifdef LOCAL
#include  "./debugger_z.cc"
#else
#define deb //
#endif

const int N = 3e5 + 5;

int solve() {
	int n;
	cin >> n;
	vector<int> a;
	a = vector<int>(n);
	rep(i , 0 , n)cin >> a[i];

	int sum = 0; FOR(i , a)sum += i;


	if (sum % n) {
		cout << -1 << endl;
		return 0;
	}

	int x = sum / n;

	int ans = 0;
	FOR(i , a) {
		if (i > x)
			ans++;
	}
	cout << ans << endl;
	return 0;

}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cout << fixed << setprecision(20);

#ifdef LOCAL
	auto start1 = chrono::high_resolution_clock::now();
	freopen("input.txt", "r", stdin);
	//freopen("output.txt","w",stdout);
#endif

	int tests = 1;
	cin >> tests;
	for (int i = 1 ; i <= tests ; i++) {
		solve();
	}

	return 0 ;
}
