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;
typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef unordered_map<int,int> umii;
typedef unordered_map<ll, ll> umll;
typedef map<int,int> mii;
typedef map<ll,ll> mll;
 
#define      unsigned long long int ull
#define      fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define      F first
#define      S second
#define      f(i,n) for (int i = 0; i < n; i++)
#define      fr(i,n) for(int i=(int)n-1;i>=0;--i)
#define      fo(i,a,n) for(ll i=(ll)a;i<(ll)n;i++)
#define      pb push_back
#define      mkp make_pair
#define      pri_que_down priority_queue<ll>
#define      pri_que_up  priority_queue<ll, vector<ll>, greater<ll> >
 void Prime(ll n)
{
    bool prime[n+1];
    memset(prime, true, sizeof(prime));
    for (int p=2; p*p<=n; p++)
    {
        if (prime[p] == true)
        {
            for (int i=p*p; i<=n; i += p)
                prime[i] = false;
        }
    }
    for (int p=2; p<=n; p++)
       if (prime[p])
          cout << p << " ";
}
void solve() 
{
   ll n,m;
   cin>>n;
   vl a(n);
   ll i;
   f(i,n)
   cin>>a[i];
   ll  sum=0;
   f(i,n)
   sum+=a[i];
   ll x=*max_element(a.begin(),a.end());
   if(sum%n==0)
   {
   	  ll p=sum/n;
   	  ll c=0;
   	  f(i,n)
   	  {
   	  	if(a[i]>p)
   	  	c++;
      }
      cout<<c<<endl;
   }
   else
   {
   	cout<<"-1"<<endl;
   }
}
int main()
{ 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin>>t;
while(t--)
{
solve();
}
return 0;
}