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
from sys import stdin, stdout, setrecursionlimit as srl
from threading import stack_size
from collections import deque, defaultdict as dd
from math import gcd, ceil

#srl(int(1e6))
input = stdin.readline
flush = stdout.flush

def ub(arr, lo, hi, val):
    # returns first index with value greater than val
    # len(arr) if all values less than equal to val
    while (lo < hi):
        mid = (lo + hi) // 2
        if (arr[mid] > val):
            hi = mid - 1
        elif (arr[mid] <= val):
            lo = mid + 1

    if (lo == len(arr) - 1 and arr[-1] <= val):
        lo = len(arr)
    
    return lo

def lb(arr, lo, hi, val):
    # returns first index with value equal to val
    # if all values less than val returns len(arr)
    # if all values greater than val returns 0
    while (lo < hi):
        mid = (lo + hi) // 2
        if (arr[mid] < val):
            lo = mid + 1
        elif (arr[mid] >= val):
            hi = mid - 1

    if (lo == len(arr) - 1 and arr[-1] < val):
        lo = len(arr)

    return lo


tc = 1
tc = int(input())
for _ in range(tc):
    n = int(input())
    a = list(map(int, input().split()))
    x = sum(a)
    if (x % n != 0):
        print('-1')
        continue
    x = x//n

    a.sort(reverse = True)
    ans = 0
    for i in a:
        if i > x:
            ans += 1
        else:
            break
    print(ans)