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 <iostream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <iterator>
#include <deque>
#include <bitset>
#include <climits>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
#include <cmath>
 
#define ll long long
//#define mod 1000000007
#define mod 998244353
#define pi pair< int,int >
#define pll pair<long long ,long long>
 
#define vi vector<int>
#define vll vector< long long  >
#define vb vector<bool>
#define vvi vector<vector<int> >
#define nl "\n"
#define mp make_pair
#define pb push_back
#define yes cout<<"YES"<<"\n"
#define no cout<<"NO"<<"\n"
#define ff first
#define ss second
 
#define for_ab(i,a,b) for(long long i=a ;i<b ;++i)
#define for_n( i , n) for_ab(i , 0ll , n)
#define all(v)  v.begin() , v.end()
 
using namespace std;
/*
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int , null_type, less_equal<int> , rb_tree_tag ,tree_order_statistics_node_update>
*/
int dir[4][2]={{0,-1} , {0,1} , {-1,0} , {1,0}};
bool ispalin(string s){
	int i=0 , j=s.size()-1;
	while(i<j){
		if(s[i]!=s[j]){
			return false;
		}
		++i;
		--j;
	}
	return true;
}
bool isprime(ll a){
	for(ll i=2;i*i<=a;++i){
		if(a%i==0){
			return false;
		}
 
	}
	return true;
}
ll ceil(ll a, ll b){
	ll c=(a+b-1)/b;
	return c;
}
 
ll gcd(ll a , ll b){
	if(a==0){
		return b;
	}
	if(b==0){
		return a;
	}
	return gcd(b,a%b);
}
ll power_mod(ll a , ll b){
	if(b==0){
		return 1;
	}
	ll temp=power_mod(a , b/2);
	temp=(temp*temp)%mod;
	if(b%2){
		temp=(temp*a)%mod;
	}
	return temp;
}
/*ll NCR(ll n , ll r){
	n=fact[n];
	
	ll a=fact[r] , b=fact[n-r];
	a=power_mod(a,mod-2);
	b=power_mod(b , mod-2);
	a=(a*b)%mod;
	n=(n*a)%mod;
	return n;
 
}*/
int main(){
		ios::sync_with_stdio(0);
		cin.tie(0);
	/*	#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif*/
		int t=1;
		cin>>t;
		
		
		
        while(t--){



        	ll n;cin>>n;
        	vll v(n);
        	ll sum=0;
        	map<ll,ll>mapi;
        	for_n(i , n){
        		cin>>v[i];
        		sum+=v[i];
        	}
        	if(sum%n){
        		cout<<-1<<nl;continue;
        	}
        	sort(all(v));
        	reverse(all(v));
        	
        	ll ans=0;
        	ll each=sum/n;
        	
        	for_n(i , n){
        		if(v[i]<=each){
        			break;
        		}
        		ans=i+1;
        	}
        	cout<<ans<<nl;
        	


        	}
        
      }	
