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
import java.io.*;
import java.util.*;
import java.math.*;

public class Solution {

    BufferedReader br;
    PrintWriter out;
    
    long mod = (long)(1e9 + 7), inf = (long)(3e18);
    
    void solve() {
        int t = ni();
        while(t-- > 0) {
            int n = ni();
            int a[] = new int[n];
            long sum = 0;
            for(int i = 0; i < n; i++) {
                a[i] = ni();
                sum += a[i];
            }
            if(sum % n != 0) {
                out.println(-1);
            } else {
                sum /= n;
                int ans = 0;
                for(int x : a)  if(x > sum) ans++;
                out.println(ans);
            }
        }
    }
    
    long mp(long b, long e, long mod) {
        long r = 1;
        while(e > 0) {
            if((e & 1) == 1)    r = (r * b) % mod;
            b = (b * b) % mod;  e >>= 1;
        }
        return r;
    }

    //---------- I/O Template ----------
    
    public static void main(String[] args) { new Solution().run(); }
    void run() { 
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    String ip[];
    int idx;
    
    String ns() {
        if(ip == null || idx == ip.length) {
            try {
                ip = br.readLine().split(" ");
                idx = 0;
            } catch (IOException e) {
                System.out.println(e);
            }
        }
        return ip[idx++];
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    char nc() {
        return ns().charAt(0);
    }
}
