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
/* ANKIT CHOUDHARY
   CSE, 3rd Year 
   IIIT UNA
 */

#include <bits/stdc++.h>
using namespace std;

#define gc getchar_unlocked
#define fastio                        \
    ios_base::sync_with_stdio(false); \
    std::cin.tie(NULL);
#define f(i, k, n) for (i = k; i < n; i++)
#define rf(i, k, n) for (i = n; i > k; i--)
#define ll long long
#define si(x) scanf("%d", &x)
#define sl(x) scanf("%lld", &x)
#define ss(s) scanf("%s", s)
#define pi(x) printf("%d\n", x)
#define pl(x) printf("%lld\n", x)
#define ps(s) printf("%s\n", s)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define clr(x) memset(x, 0, sizeof(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
ll MOD = 1000000007;

template <class T>
T power(T base, T num)
{
    ll res = 1;
    while (num > 0)
    {
        if (num & 1ll)
            res = res * base;

        num >>= 1ll;
        base = base * base;
    }
    return res;
}

template <class T>
T modPower(T base, T num)
{
    ll res = 1;
    while (num > 0)
    {
        if (num & 1ll)
            res = res * base % MOD;

        num >>= 1ll;
        base = base * base % MOD;
    }
    return res % MOD;
}

template <class T>
T modinverse(T a)
{
    return power(a, MOD - 2);
}

ll gcd(ll a, ll b)

{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int32_t main()
{
    fastio;
    int t = 1;
    cin >> t;
    while (t--)
    {
        ll n, k, temp;
        cin >> n;
        int arr[n];

        int count;
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            cin >> arr[i];
            sum += arr[i];
        }
        if (sum % n)
        {
            cout << -1 << "\n";
            continue;
        }

        sum /= n;
        int ans = 0;
        for (int i = 0; i < n; i++)
            if (arr[i] > sum)
                ans++;

        cout << ans << "\n";
    }
    return 0;
}