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<math.h>
#include<string>
#include<algorithm>
#include<bits/stdc++.h>
#include<bits/stdc++.h>
#define ll long long int
#define ull unsigned long long int
#define MOD 1000000007
#define f(i,a,b) for(int i=a;i<b;i++)
#define fe(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=b;i>a;i--)
#define fde(i,a,b) for(int i=b;i>=a;i--)
#define fastio cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(false);
#define debug(x) cout << '>' << #x << ':' << x << endl;
#define F first
#define S second
#define mp make_pair
#define epsilon 1e-15
#define pb push_back
#define PI 3.1415926535897932384626433832
using namespace std;
ll power(ll a,ll b,ll mod)
{
        ll res=1;
        a%=mod;
        while(b>0)
        {
                if(b&1)
                res=(res*a)%mod;

                a=(a*a)%mod;
                b>>=1;

        }
        return res;
}

ll fact[200001]={0};
ll invfact[200001]={0};

ll facto(ll mod)
{
        fact[0]=1;
        fact[1]=1;
        for(int i=2;i<=200000;++i)
        fact[i]=(i*fact[i-1])%mod;
        
        for(int i=0;i<=200000;++i)
        {
                invfact[i]=power(fact[i],mod-2,mod);
        }
}
ll ncr(ll a,ll b,ll mod)
{
        if(a<b)
        return 0;

        

        ll ans=1;
        ans=(ans*fact[a]);
        ans=(ans*invfact[a-b])%mod;
        ans=(ans*invfact[b])%mod;

        return ans;
   
}


int main()
{
        int t;
        cin>>t;
        while(t--)
        {
                int n;
                cin>>n;
                int arr[n];
                for(int i=0;i<n;++i)
                {
                        cin>>arr[i];
                        
                }
                
                int cnt=0;
                ll sum=0;
                for(auto &i:arr)
                sum+=i;

                if(sum%n!=0)
                cout<<"-1\n";
                else
                {
                        int maxi=*max_element(arr,arr+n);
                        
                        
                                for(auto &i:arr)
                                {
                                        if(i>(sum/n))
                                        cnt++;
                                }
                                cout<<cnt<<"\n";
                        
                }


        }



        
        
}