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>
/*
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
*/
using namespace std;
/*
typedef tree<int,null_type, less<int>,rb_tree_tag,
        tree_order_statistics_node_update> indexed_set;
*/
#define pb push_back
#define ll long long
#define x1 first
#define y1 second
#define frb(i,a,b) for(ll (i)=(a);(i) <= (b); (i)+=(i)&-(i))
#define rfrb(i,a) for(ll (i)=(a);(i) > 0;(i)-=(i)&-(i))
#define fr(i,a,b) for(ll (i) = (a); (i) <= (b); (i)++)
#define rep(i,c) for(auto (i) : (c))
#define mini(a,b) (((a) < (b)) ? (a) : (b))
#define maxi(a,b) (((a) > (b)) ? (a) : (b))
#define par pair<ll,ll>
#define vp vector<par>
#define vi vector<ll>
#define dq deque<ll>
//indexed_set s; 
///always see the properties of the problems
///siempre poner ("\n" cout.flush) como endl
///----------------------------------------------------------------------------

const ll mod = 1e9+7;
const ll INF = 1e18;
ll n,t;
ll ar[200005];     

int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);
    cout.setf(ios::fixed);cout.precision(0);
    ///freopen("a.in","r",stdin);
    ///freopen("a.out","w",stdout);
    cin>>t;
    while(t--){
        cin>>n;
        ll cnt = 0;
        fr(i,1,n){
            cin>>ar[i];
            cnt+=ar[i];
        }
        if(cnt%n!=0){
            cout<<-1<<"\n";
            continue;
        }
        cnt = cnt/n;
        ll sol = 0;
        fr(i,1,n){
            if(ar[i]>cnt)sol++;
        }
        cout<<sol<<"\n";
    }
    return 0;
}