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 ll;
typedef long double lld;
#define  loop(n)        for(int i=0;i<(n);++i)
#define  rep(i,zz,n)    for(int i=zz;i<(n);++i)
#define  all(v)         v.begin(),v.end()
#define  pb             push_back
#define  ff             first
#define  ss             second
#define  sz(x)          (int)x.size()
#define  nline          cout<<'\n'

#define  debug(x)       cout<<#x<<" ";_print(x);cout<<'\n';
#define  _debug(a,b)    cout<<#a<<" [ ";array_debug(a,b);cout<<"]\n";
void array_debug(ll *a, ll n){ for(ll i=0;i<(n);++i) cout<<a[i]<<' ';}
void array_debug(int *a, int n){ for(int i=0;i<(n);++i) cout<<a[i]<<' ';}
void array_debug(ll *a, int n){ for(int i=0;i<(n);++i) cout<<a[i]<<' ';}
void _print(ll t) {cout << t;}
void _print(int t) {cout << t;}
void _print(string t) {cout << t;}
void _print(char t) {cout << t;}
void _print(lld t) {cout << t;}
void _print(double t) {cout << t;}
template <class T, class V> void _print(pair <T, V> p) {cout << "{"; _print(p.ff); cout << ","; _print(p.ss); cout << "}";}
template <class T> void _print(vector <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(set <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(multiset <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(map <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}

const int mod=1e9+7;
const ll inf=1e18;
const int N1=2e5+10;

int code(){
    int n;
    cin>>n;
    int a[n],sum=0;
    loop(n){
        cin>>a[i];
        sum+=a[i];
    }
    if(sum%n){
        cout<<-1;
        nline;
        return 0;
    }
    int ans=0;
    loop(n){
        if(a[i]>sum/n)  ++ans;
    }
    cout<<ans;
    nline;
    return 0;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    int tc=1;
    cin>>tc;
    loop(tc)  code();
    return 0;
}