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>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
using namespace std;
#define ll long long int
#define ld long double
#define mod (ll)(1e9+7)
#define for0(i,n) for(ll i=0;i<n;i++)
#define for1(i,n) for(ll i=1;i<=n;i++)
#define forab(i,a,b) for(ll i=a;i<=b;i++)
#define vll vector<ll>
#define pb(i) push_back(i)
#define fio ios_base::sync_with_stdio(false),cin.tie(NULL)
#define maxheap priority_queue<ll>
#define minheap priority_queue<ll,vector<ll>,greater<ll>>

const ll inf = 1e9 + 7;

ll mult(ll a, ll b, ll p) {return ((a % p) * (b % p)) % p;}
ll add(ll a, ll b, ll p) {return (a % p + b % p) % p;}
ll fpow(ll n, ll k, ll p) {
	ll r = 1;
	for (; k; k >>= 1) {
		if (k & 1) r = r * n % p;
		n = n * n % p;
	}
	return r;
}
ll gcd(ll a, ll b) {if (a == 0) return b; return gcd(b % a, a);}
ll max(ll a, ll b) {if (a > b) return a; else return b;}
ll min(ll a, ll b) {if (a < b) return a; else return b;}

// Code

void solve() {

	ll n;
	cin >> n;

	vll a(n);
	ll sum = 0;

	for0(i, n) {
		cin >> a[i];
		sum += a[i];
	}

	if (sum % n != 0) {
		cout << "-1\n";
		return;
	}

	sum = sum / n;

	ll count = 0;

	for0(i, n) {
		if (a[i] > sum) count++;
	}

	cout << count << "\n";

}

int main()
{

	fio;

	ll t = 1;
	cin >> t;

	while (t--) {

		solve();

	}

	return 0;

}