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
#include <cstring>
#include <string>
#include <time.h>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <functional>
#include <limits.h>
#include <bitset>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <iostream>
#include <stdio.h>
#include <fstream>
#include <random>
using namespace std;
#pragma warning(disable : 4996)
typedef long double ld;
typedef double db;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<double, double> pd;
#define gets(x) fgets(x, sizeof(x), stdin)
#define FIO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define For(i, a, b) for(ll i=a; i<b; i++)
#define inc(i, a, b) for(ll i=a; i<=b; i++)
#define dec(i, a, b) for(ll i=a; i>=b; i--) 
#define pb push_back
#define sz(a) ((int)(a.size()))
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define ZERO(a) memset(a,0,sizeof(a))
#define X first
#define Y second
#define ub upper_bound
#define lb lower_bound
#define minh priority_queue <ll, vector<ll>, greater<ll>>
#define minph priority_queue <pl, vector<pl>, greater<pl>>
#define maxh priority_queue <ll>
#define M_PI 3.14159265358979323846
#define eps (double)(0.000000001)
const ll N = 1000007, M = 1e7, md = 1e9 + 7, inf = 1e9 + 5, linf = 2e18 + 5, MX = 3e5, aA = 'a' - 'A';
stack <int> stk;
ll a[N], b[N], p[N], w, h, x, qry, y, z, l, r, cnt, res, tmp, num, n, m, k, T;
string s;
vi g[N]; bool c[N];

int main() {
	FIO;
	cin >> T;
	while (T--) {
		cin >> n;
		y = cnt = 0;
		For(i, 0, n) {
			cin >> a[i];
			y += a[i];
		}
		if (y%n) cout << -1 << '\n';
		else {
			k = y / n;
			For(i, 0, n)
				if (a[i] > k) cnt++;

			cout << cnt << '\n';
		}
	}
}