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
//package Div_2B_Problems;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

/*Backtracking is a general algorithm for finding all (or some) solutions to some computational problems, notably constraint satisfaction problems, that incrementally builds candidates to the solutions, and abandons a candidate ("backtracks") as soon as it determines that the candidate cannot possibly be completed to a valid solution.[1]

The classic textbook example of the use of backtracking is the eight queens puzzle, that asks for all arrangements of eight chess queens on a standard chessboard so that no queen attacks any other. In the common backtracking approach, the partial candidates are arrangements of k queens in the first k rows of the board, all in different rows and columns. Any partial solution that contains two mutually attacking queens can be abandoned.

Backtracking can be applied only for problems which admit the concept of a "partial candidate solution" and a relatively quick test of whether it can possibly be completed to a valid solution. It is useless, for example, for locating a given value in an unordered table. When it is applicable, however, backtracking is often much faster than brute-force enumeration of all complete candidates, since it can eliminate many candidates with a single test.

Backtracking is an important tool for solving constraint satisfaction problems,[2] such as crosswords, verbal arithmetic, Sudoku, and many other puzzles. It is often the most convenient technique for parsing,[3] for the knapsack problem and other combinatorial optimization problems. It is also the basis of the so-called logic programming languages such as Icon, Planner and Prolog.

Backtracking depends on user-given "black box procedures" that define the problem to be solved, the nature of the partial candidates, and how they are extended into complete candidates. It is therefore a metaheuristic rather than a specific algorithm – although, unlike many other meta-heuristics, it is guaranteed to find all solutions to a finite problem in a bounded amount of time.*/
public class Codeforces725_B {
	public static void main(String args[]) {
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		while(t>0) {
			int n=fs.nextInt();
			int []arr=new int[n];
			long sum=0;
			int min=Integer.MAX_VALUE;
			int max=Integer.MIN_VALUE;
			for(int i=0;i<n;i++) {
				arr[i]=fs.nextInt();
				sum+=arr[i];
				min=Math.min(min, arr[i]);
				max=Math.max(max, arr[i]);
			}
			if(sum%n!=0)
				System.out.println(-1);
			else {
				Arrays.sort(arr);
				long avg=sum/n;
				//System.out.println(avg);
				long diff=0;
				for(int i=0;i<n;i++) {
					if(arr[i]<avg)
						diff+=Math.abs(avg-arr[i]);
					else break;
				}
				long sum2=0;
				int cnt=0;
				//System.out.println(diff);
				for(int i=arr.length-1;i>=0;i--) {
					if(arr[i]>avg) {
						sum2+=(arr[i]-avg);
						cnt++;}
					if(sum2>=diff)
						break;
				}
				System.out.println(cnt);
			}
			
			t--;
		}
	}
	private static int[] readArray(int n) {
		// TODO Auto-generated method stub
		return null;
	}

static final Random random=new Random();
	
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
}
}
