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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class Solve {

	static int mod = 1000000000 + 7;
	static long INF = 1000_000_000_000_000L + 100;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);

		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int a[] = sc.nextIntArray(n);
			long sum = 0;
			for (int i : a)
				sum += i;

			if (sum % n != 0) {
				pw.println(-1);
			} else {
				long l = sum / n;
				int ans = 0;
				for (int i : a)
					if (i > l)
						ans++;

				pw.println(ans);

			}

		}

		pw.flush();

	}

	static public boolean[] sieve(int n) {
		boolean prime[] = new boolean[n + 1];
		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;

	}

	static boolean isPalindrome(String str) {

		// Pointers pointing to the beginning
		// and the end of the string
		int i = 0, j = str.length() - 1;

		// While there are characters to compare
		while (i < j) {

			// If there is a mismatch
			if (str.charAt(i) != str.charAt(j))
				return false;

			// Increment first pointer and
			// decrement the other
			i++;
			j--;
		}

		// Given string is a palindrome
		return true;
	}

	public static int primeFactors(int n) {
		int count = 0;
		// Print the number of 2s that divide n
		while (n % 2 == 0) {
			n /= 2;
		}

		if (n == 1)
			return 0;

		// n must be odd at this point. So we can
		// skip one element (Note i = i +2)
		for (int i = 3; i <= Math.sqrt(n); i += 2) {
			// While i divides n, print i and divide n
			while (n % i == 0) {
				n /= i;
				count++;
			}
		}

		if (n > 1)
			count++;

		return count;

	}

	static int gcd(int a, int b) {
		if (a == 0)
			return b;
		if (b == 0)
			return a;

		return gcd(b, a % b);

	}

	public static int[] swap(int data[], int left, int right) {

		// Swap the data
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;

		// Return the updated array
		return data;
	}

	static long nCrModp(int n, int r, int p) {
		if (r > n - r)
			r = n - r;

		// The array C is going to store last
		// row of pascal triangle at the end.
		// And last entry of last row is nCr
		int C[] = new int[r + 1];

		C[0] = 1; // Top row of Pascal Triangle

		// One by constructs remaining rows of Pascal
		// Triangle from top to bottom
		for (int i = 1; i <= n; i++) {

			// Fill entries of current row using previous
			// row values
			for (int j = Math.min(i, r); j > 0; j--)

				// nCj = (n-1)Cj + (n-1)C(j-1);
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r];
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

}

class Pair implements Comparable<Pair> {
	int tower;
	int value = 0;

	Pair(int x, int y) {
		tower = x;
		value = y;
	}

	public int compareTo(Pair o) {
		return this.value - o.value;
	}

}

class Scanner {
	BufferedReader br;
	StringTokenizer st;

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

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

	int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(next());
	}

	int[] nextIntArray(int n) throws NumberFormatException, IOException {
		int a[] = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = Integer.parseInt(next());
		return a;
	}

	long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(next());
	}

	long[] nextLongArray(int n) throws NumberFormatException, IOException {
		long a[] = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = Long.parseLong(next());
		return a;
	}

	double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(next());
	}
}