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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.security.Key;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;

public class B {

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

        char[] readCharArray(int n) {
            char[] a = new char[n];
            for (int i = 0; i < n; i++)
                a[i] = next().charAt(0);
            return a;
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static Boolean[] sieveOfEratosthenes(int n, Boolean[] prime) {
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true) {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;

        // Print all prime numbers
    }

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static void main(String[] args) {
        FastReader fr = new FastReader();

        int t = fr.nextInt();
        while (t-- > 0) {
            int n = fr.nextInt();
            int a[] = fr.readIntArray(n);
            long sum = 0l;
            for (Integer i : a) {
                sum += i;
            }
            if (sum % n != 0) {
                System.out.println(-1);
            } else {
                int count = 0;
                for (Integer i : a) {
                    if (i > (sum / n))
                        count++;
                }
                count = (1 == n) ? 0 : count;
                System.out.println(count);
            }

        }
    }

}
