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 __future__ import division, print_function
import bisect
import math
import heapq
import itertools
import sys
from collections import deque
from atexit import register
from collections import Counter
from functools import reduce
sys.setrecursionlimit(100000)
if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream
 
 
if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
 
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)
 
        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)
 
        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)
 
    input = raw_input
    range = xrange
 
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
 
 
def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.
 
    Args:
        sync (bool, optional): The new synchronization setting.
 
    """
    global input, flush
 
    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        def input(): return sys.stdin.readline().rstrip('\r\n')
 
        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
 
 
def dd():
    return map(int, input().split())
 
 
def arr():
    return list(map(int, input().split()))
 
 
def twinSort(X, Y):
    # sorting X wrt Y
    return [x for _, x in sorted(zip(Y, X))]
 
def check1(rem,d2):
    while rem:
        d=rem%10
        if d==d2:
            return True
        rem//=10
    return False
 
def solve():
    n=int(input())
    ar=arr()
    if math.ceil(sum(ar)/n)!=math.floor(sum(ar)/n):
        print(-1)
    else:
        avg=sum(ar)//n
        k=0
        for i in ar:
            if i>avg:
                k+=1
        print(k)
 
 
def main():
    testCase = 1
    if testCase:
        for _ in range(int(input())):
            solve()
    else:
        solve()
 
 
if __name__ == '__main__':
    sync_with_stdio(False)
    main()