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 <vector>
 #include <queue>
#include <unordered_map>
 #include <climits>
 #include <algorithm>
 #include <cmath>
#include <bitset>
 #include <cstdio>
 #include <cstring>
 #include <stack>
 #include <iomanip>
 #include <set>
#include <map>
using namespace std;
#define F                    first
#define S                    second
#define int                  long long int
 #define pb                   push_back
 #define pii                  pair<int,int>
#define vi                   vector<int>
#define vii                  vector <pair <int,int> >
#define umpii                unordered_map<int,int>
#define mem(a,b)             memset(a, b, sizeof a)
#define mii                  map<int,int>
#define m_p                  make_pair
#define pqh                  priority_queue<int>
#define pql                  priority_queue<int,vi,greater <int> >
#define setbits(x)           __builtin_popcountll(x);
#define zerobits(x)          __builtin_ctzll(x);
#define fo(i, k, n)          for (int i = k; k < n ? i < n : i > n; k < n ? i++ : i--)
#define fa(i, N)             for (auto i : N)
#define all(X)               X.begin(), X.end()
#define ceill(a,b)           ((a)/(b) + (((a)%(b))!=0))
#define bin(n)               bitset<32>(n).to_string()
#define yes                  cout<<"YES";

#define no                   cout<<"NO";

#define vin(name,N)          vi name(N);fo(i,0,N){cin>>name[i];}
#define dbg(x)               cerr<<#x<<" :: "<<x<<endl;
 #define dbg2(x,y)            cerr<<#x<<" :: "<<x<<"	"<<#y<<" :: "<<y<<endl;
const int INF = 1e9;
#define mod 1000000007
void solve()
{
   int n;
   cin>>n;
   int arr[n];
   int sum = 0,count = 0;
   for(int i=0;i<n;i++){
       cin>>arr[i];
       sum+=arr[i];
   }
   if(sum%n == 0){
       int x = sum / n;
       for(int i=0;i<n;i++)if(arr[i]> x)count++;
       cout<<count<<endl;
   }
   else cout<<-1<<endl;
   
}
int32_t main()
{
int t;
cin>>t;
while(t--)
{
    solve();
}
 return 0;
 }