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://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
#define pfi(a) printf("%d ",a);
#define pfl(a) printf("%lld",a);
#define pfin(a) printf("%d\n",a);
#define pfln(a) printf("%lld\n",a);
#define sfi(a) scanf("%d",&a);
#define sfl(a) scanf("%lld",&a);
#define f(i,a,b) for(ll i=a;i<b;i++)
#define fb(i,b,a) for(ll i=b-1;i>=a;i--)
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ss second
#define ff first
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a);
#define tc ll t;sfl(t);while(t--)
#define ll long long
#define mem(a,f) memset(a,false,sizeof(a));
#define sor(v) sort(v.begin(),v.end())
#define vi vector<int>
#define vc vector<char>
#define vl vector<ll>
#define endl "\n"
int main()
{
 tc
 {
 	int n;
 	sfi(n);
 	int cnt=0,sum=0,ar[n];
 	f(i,0,n)
 	{
 	  sfi(ar[i]);
 	  sum+=ar[i];
 	  if(i>0)
 	  {
 	  	if(ar[i]==ar[i-1])
 	  	{
 	  		cnt++;
 	  	}
 	  }	
 	}
 	if(cnt==(n-1))
 	{
 		pfin(0);
 	}
 	else 
 	{
 		if(sum%n!=0)
 		{
 			pfin(-1);
 		}
 		else
 		{
 			int x=sum/n;
 			int ans=0;
 			f(i,0,n)
 			{
 				if(ar[i]>x)
 				{
 					ans++;
 				}
 			}
 			pfin(ans);
 		}
 	}
 }
 return 0;
}