#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <stack>
#include <set>
#include <queue>
#include <string>
#include <deque>
#include <iomanip>
#include <map>
#include <ctime>  

using namespace std;

typedef long long ll;

ll bin_pow(ll a, ll n) {
	if (n == 0)
		return 1;

	if (n % 2 == 0) {
		ll b = bin_pow(a, n / 2);
		return b * b;
	}
	else return bin_pow(a, n - 1) * a;
}
ll gcd(ll a, ll b) {
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

vector<ll>arr;
vector<ll>pref;

void create_pref() {
	pref.resize(arr.size() + 1);
	pref[0] = 0;
	for (int i = 1; i < pref.size(); i++) {
		pref[i] = pref[i - 1] + arr[i - 1];
	}
}

ll ans1(int i) {
	return pref[i];
}

ll ans2(int i) {
	return (pref.back() - pref[i + 1]) - (arr[i] * (arr.size() - i - 1));
}

int main() {
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		map<int, ll>m;
		for (int i = 0; i < n; i++) {
			int a; cin >> a;
			m[a]++;
		}
		for (auto i : m)
			arr.push_back(i.second);
		sort(arr.begin(), arr.end());
		create_pref();
		ll ans = 1e18;
		for (int i = 0; i < arr.size(); i++) {
			ans = min(ans, ans1(i) + ans2(i));
		}
		cout << ans << '\n';
		arr.clear();
		pref.clear();
	}
	return 0;
}
### Prompt

In CPP, your task is to solve the following problem:

Polycarp was gifted an array a of length n. Polycarp considers an array beautiful if there exists a number C, such that each number in the array occurs either zero or C times. Polycarp wants to remove some elements from the array a to make it beautiful.

For example, if n=6 and a = [1, 3, 2, 1, 4, 2], then the following options are possible to make the array a array beautiful: 

  * Polycarp removes elements at positions 2 and 5, array a becomes equal to [1, 2, 1, 2]; 
  * Polycarp removes elements at positions 1 and 6, array a becomes equal to [3, 2, 1, 4]; 
  * Polycarp removes elements at positions 1, 2 and 6, array a becomes equal to [2, 1, 4]; 



Help Polycarp determine the minimum number of elements to remove from the array a to make it beautiful.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case consists of one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output one integer — the minimum number of elements that Polycarp has to remove from the array a to make it beautiful.

Example

Input


3
6
1 3 2 1 4 2
4
100 100 4 100
8
1 2 3 3 3 2 6 6


Output


2
1
2

### Response

```cpp
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <stack>
#include <set>
#include <queue>
#include <string>
#include <deque>
#include <iomanip>
#include <map>
#include <ctime>  

using namespace std;

typedef long long ll;

ll bin_pow(ll a, ll n) {
	if (n == 0)
		return 1;

	if (n % 2 == 0) {
		ll b = bin_pow(a, n / 2);
		return b * b;
	}
	else return bin_pow(a, n - 1) * a;
}
ll gcd(ll a, ll b) {
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

vector<ll>arr;
vector<ll>pref;

void create_pref() {
	pref.resize(arr.size() + 1);
	pref[0] = 0;
	for (int i = 1; i < pref.size(); i++) {
		pref[i] = pref[i - 1] + arr[i - 1];
	}
}

ll ans1(int i) {
	return pref[i];
}

ll ans2(int i) {
	return (pref.back() - pref[i + 1]) - (arr[i] * (arr.size() - i - 1));
}

int main() {
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		map<int, ll>m;
		for (int i = 0; i < n; i++) {
			int a; cin >> a;
			m[a]++;
		}
		for (auto i : m)
			arr.push_back(i.second);
		sort(arr.begin(), arr.end());
		create_pref();
		ll ans = 1e18;
		for (int i = 0; i < arr.size(); i++) {
			ans = min(ans, ans1(i) + ans2(i));
		}
		cout << ans << '\n';
		arr.clear();
		pref.clear();
	}
	return 0;
}
```