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

// Problem: B. Friends and Candies
// Contest: Codeforces - Codeforces Round #725 (Div. 3)
// URL: https://codeforces.com/contest/1538/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define ff first
#define ss second
#define VI vector <int>
#define VVI vector <VI>
#define pii pair<int, int>
#define ppi pair<pii, int>
#define mii map<int, int>
#define mci map<char, int>
#define miv map<int, VI>
#define mis map<int, set<int>>
#define setbits(n) __builtin_popcount(n)
#define all(v) (v).begin(), (v).end()
#define yes cout<<"Yes"<<endl
#define no cout<<"No"<<endl
#define endl "\n"
#define show2(a, b) cout<<a<<' '<<b<<endl
#define show3(a, b, c) cout<<a<<' '<<b<<' '<<c<<endl
#define show(arr) for (auto i:arr) cout<<i<<' '; 

const long long N=1e5+5;
const long long mod=1000000007; //998244353

void cases(){
    int n;
    cin>>n;
    int sum=0;
    mii m;
    for (int i=0; i<n; i++){
    	int x;
    	cin>>x;
    	sum+=x;
    	m[x]++;
    }
    if (sum%n){
    	cout<<-1<<endl;
    	return;
    }
    sum/=n;
    int ans=0;
    for (auto i:m){
    	if (i.ff>sum) ans+=i.ss;
    }
    cout<<ans<<endl;
}

int32_t main(){
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    int t=1;
    cin>>t;
    for (int i=0; i<t; i++){
    	//cout<<"Case #"<<i+1<<": ";
		cases();
    }
    return 0;
}
