#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define ll long long
void solve()
{
    int n;
    cin >> n;
    ll a[n];
	ll hmany[n + 1];
    for(int i = 0; i < n; ++i)
    {
    	cin >> a[i];
    	hmany[i] = 1;
    }
    sort(a, a + n);
    hmany[n] = 1;
    hmany[0] = 1;
    vector <ll> vec;
    int cnt = 0;
    for(int i = 1; i < n + 1; ++i)
    {
    	if(a[i] == a[i - 1])
    	{
    		hmany[i] = hmany[i - 1] + 1;
    	}
    	else
    	{
    		vec.push_back(hmany[i - 1]);
    		hmany[i] = 1;
    	}
    }
    
    vector <ll> fin;
    sort(vec.begin(), vec.end());
    
    for(int i = 0; i < n; ++i)
    {
    //	cout << hmany[i] << " ";
    }
    //cout << endl;
    ll s = 0;
    for(int i = 0; i < vec.size(); ++i)
    {
    	s += vec[i];
    	//cout << vec[i] << " ";
    }
    //cout << endl;
    //cout << endl;
    
    //ll s = 0;
    /*
    for(int i = 0; i < vec.size(); ++i)
    {
    	if(i == 0)
    	{
    		fin.push_back(vec[i]);
    		//s += vec[i];
    	}
    	else
    	{
    		if(vec[i] != fin.back())
    		{
    			fin.push_back(vec[i]);
    			//s += vec[i];
    		}
    	}
    	s += vec[i];
    }
    */
    ll min = s - vec[vec.size() - 1];
    ll diff = 0;
    ll mult = 1;
    ll sback = vec[vec.size() - 1];
    for(int i = vec.size() - 2; i >= 0; i--)
    {
    	diff += (vec.size() - i - 1) * (vec[i + 1] - vec[i]);
		if(diff == 0)
		{
			 //mult++; 
		}
		else
		{
			//diff *= mult;
			//mult = 1;
		}
    	sback += vec[i];
    	if(s - sback + diff < min)
    	{
    		min = s - sback + diff;
    	}
    	//cout << min << " ";
    	
    }
    cout << min << endl;
    /*
    for(int i = 0; i < fin.size(); ++i)
    {
    	cout << fin[i] << " ";
    }
    cout << endl;
    */
}
int main() {
	int q;
	cin >> q;
	while(q--)
	{
	    solve();
	}
	return 0;
}


### Prompt

Please provide a CPP coded solution to the problem described below:

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 <algorithm>
#include <vector>
using namespace std;
#define ll long long
void solve()
{
    int n;
    cin >> n;
    ll a[n];
	ll hmany[n + 1];
    for(int i = 0; i < n; ++i)
    {
    	cin >> a[i];
    	hmany[i] = 1;
    }
    sort(a, a + n);
    hmany[n] = 1;
    hmany[0] = 1;
    vector <ll> vec;
    int cnt = 0;
    for(int i = 1; i < n + 1; ++i)
    {
    	if(a[i] == a[i - 1])
    	{
    		hmany[i] = hmany[i - 1] + 1;
    	}
    	else
    	{
    		vec.push_back(hmany[i - 1]);
    		hmany[i] = 1;
    	}
    }
    
    vector <ll> fin;
    sort(vec.begin(), vec.end());
    
    for(int i = 0; i < n; ++i)
    {
    //	cout << hmany[i] << " ";
    }
    //cout << endl;
    ll s = 0;
    for(int i = 0; i < vec.size(); ++i)
    {
    	s += vec[i];
    	//cout << vec[i] << " ";
    }
    //cout << endl;
    //cout << endl;
    
    //ll s = 0;
    /*
    for(int i = 0; i < vec.size(); ++i)
    {
    	if(i == 0)
    	{
    		fin.push_back(vec[i]);
    		//s += vec[i];
    	}
    	else
    	{
    		if(vec[i] != fin.back())
    		{
    			fin.push_back(vec[i]);
    			//s += vec[i];
    		}
    	}
    	s += vec[i];
    }
    */
    ll min = s - vec[vec.size() - 1];
    ll diff = 0;
    ll mult = 1;
    ll sback = vec[vec.size() - 1];
    for(int i = vec.size() - 2; i >= 0; i--)
    {
    	diff += (vec.size() - i - 1) * (vec[i + 1] - vec[i]);
		if(diff == 0)
		{
			 //mult++; 
		}
		else
		{
			//diff *= mult;
			//mult = 1;
		}
    	sback += vec[i];
    	if(s - sback + diff < min)
    	{
    		min = s - sback + diff;
    	}
    	//cout << min << " ";
    	
    }
    cout << min << endl;
    /*
    for(int i = 0; i < fin.size(); ++i)
    {
    	cout << fin[i] << " ";
    }
    cout << endl;
    */
}
int main() {
	int q;
	cin >> q;
	while(q--)
	{
	    solve();
	}
	return 0;
}


```