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
/* 
    Author: Aman Patel
    Date: 10-06-2021
*/


import java.io.*;
import java.util.StringTokenizer;

public class CF725_DIV3 {

    public static void main(String[] args) {

        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int t;
        t = fs.nextInt();
        while (t-- > 0) {

            int n;
            n = fs.nextInt();

            long sum = 0;
            int[] nums = new int[n];
            for (int i = 0; i < n; i++) {
                nums[i] = fs.nextInt();
                sum += nums[i];
            }

            if (check(n, sum)) {
                long div = (sum / n);
                long result = 0;
                for (int i = 0; i < n; i++) {
                    if (nums[i] > div)
                        result++;
                }

                out.println(result);
            } else
                out.println(-1);
        }

        out.close();
    }

    static boolean check(int n, long sum) {
        return Math.ceil(sum / (n * 1.)) == Math.floor(sum / (n * 1.));
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }
    }

    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
}
