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

int32_t main() {

	ios::sync_with_stdio(false);
	cin.tie(nullptr);

#ifndef ONLINE_JUDGE
	freopen("input.in", "r", stdin);
	freopen("output.in", "w", stdout);
#endif

	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		int a[n];
		map<int, int> m;
		for (int i = 0; i < n; i++) {
			cin >> a[i];
			m[a[i]]++;
		}
		set<int> s;
		map<int, int> o;
		for (auto i : m) {
			s.insert(i.second);
			o[i.second]++;
		}
		vector<int> v;
		for (auto i : s) {
			v.push_back(i);
		}
		int sz = v.size();
		int pre[sz], suf[sz];
		pre[0] = v[0] * o[v[0]];
		for (int i = 1; i < sz; i++) {
			pre[i] = pre[i - 1] + v[i] * o[v[i]];
		}

		suf[sz - 1] = o[v[sz - 1]];
		for (int i = sz - 2; i >= 0; i--) {
			suf[i] = suf[i + 1] + o[v[i]];
		}
		int ssuf[sz];
		ssuf[sz - 1] = 0;
		for (int i = sz - 2; i >= 0; i--) {
			ssuf[i] = ssuf[i + 1] + suf[i + 1] * (v[i + 1] - v[i]);
		}
		int best = n;
		for (int i = 0; i < sz; i++) {
			int cur;
			if (i == 0) {
				cur = ssuf[i];
			}
			else {
				cur = pre[i - 1] + ssuf[i];
			}
			best = min(best, cur);
		}
		cout << best << "\n";
	}

	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<bits/stdc++.h>
using namespace std;
#define int long long

int32_t main() {

	ios::sync_with_stdio(false);
	cin.tie(nullptr);

#ifndef ONLINE_JUDGE
	freopen("input.in", "r", stdin);
	freopen("output.in", "w", stdout);
#endif

	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		int a[n];
		map<int, int> m;
		for (int i = 0; i < n; i++) {
			cin >> a[i];
			m[a[i]]++;
		}
		set<int> s;
		map<int, int> o;
		for (auto i : m) {
			s.insert(i.second);
			o[i.second]++;
		}
		vector<int> v;
		for (auto i : s) {
			v.push_back(i);
		}
		int sz = v.size();
		int pre[sz], suf[sz];
		pre[0] = v[0] * o[v[0]];
		for (int i = 1; i < sz; i++) {
			pre[i] = pre[i - 1] + v[i] * o[v[i]];
		}

		suf[sz - 1] = o[v[sz - 1]];
		for (int i = sz - 2; i >= 0; i--) {
			suf[i] = suf[i + 1] + o[v[i]];
		}
		int ssuf[sz];
		ssuf[sz - 1] = 0;
		for (int i = sz - 2; i >= 0; i--) {
			ssuf[i] = ssuf[i + 1] + suf[i + 1] * (v[i + 1] - v[i]);
		}
		int best = n;
		for (int i = 0; i < sz; i++) {
			int cur;
			if (i == 0) {
				cur = ssuf[i];
			}
			else {
				cur = pre[i - 1] + ssuf[i];
			}
			best = min(best, cur);
		}
		cout << best << "\n";
	}

	return 0;
}
```