#include <bits/stdc++.h>
#define forr(i,a,b) for(int i=(a);i<(b);i++)
#define forn(i,n) forr(i,0,n)
#define dforn(i,n) for(int i=n-1;i>=0;i--)
#define forall(it,v) for(auto it=v.begin();it!=v.end();it++)
#define sz(c) ((int)c.size())
#define rsz resize
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fst first
#define snd second

#ifdef ANARAP
//prueba
#else
//real
#endif

using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

int main()
{
	#ifdef ANARAP
		freopen("input.in", "r", stdin);
		//freopen("output.out", "w", stdout);
		unsigned t0, t1; t0=clock();
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int t; cin >> t;
	forn(T,t){
		int n; cin>>n;
		vector<int> a(n);
		map<int,int> frec;
		forn(i,n) cin>>a[i];
		forn(i,n) frec[a[i]]++;
		vector<int> b;
		forall(it,frec) b.pb(it->snd);
		sort(b.begin(), b.end());
		int tb=sz(b);
		vector<int> acum(tb);
		acum[0]=b[0];
		forr(i,1,tb) acum[i]=acum[i-1]+b[i];
		//~ forn(i,tb) cout << b[i] << " ";
		//~ cout << endl;
		int cant=0;
		int suma=0;
		int ans=n;
		dforn(i,tb){
			int auxacum=0;
			if(i) auxacum=acum[i-1];
			//~ cout << auxacum << " " << suma-b[i]*cant << endl;
			int pans = auxacum+(suma-b[i]*cant);
			cant++;
			suma+=b[i];
			ans=min(pans,ans);
		}
		cout << ans << '\n';
	}
	
	#ifdef ANARAP
	t1 = clock();double time = (double(t1-t0)/CLOCKS_PER_SEC); cout << "Execution Time: " << time << endl;
	#endif
	return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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>
#define forr(i,a,b) for(int i=(a);i<(b);i++)
#define forn(i,n) forr(i,0,n)
#define dforn(i,n) for(int i=n-1;i>=0;i--)
#define forall(it,v) for(auto it=v.begin();it!=v.end();it++)
#define sz(c) ((int)c.size())
#define rsz resize
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define fst first
#define snd second

#ifdef ANARAP
//prueba
#else
//real
#endif

using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

int main()
{
	#ifdef ANARAP
		freopen("input.in", "r", stdin);
		//freopen("output.out", "w", stdout);
		unsigned t0, t1; t0=clock();
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int t; cin >> t;
	forn(T,t){
		int n; cin>>n;
		vector<int> a(n);
		map<int,int> frec;
		forn(i,n) cin>>a[i];
		forn(i,n) frec[a[i]]++;
		vector<int> b;
		forall(it,frec) b.pb(it->snd);
		sort(b.begin(), b.end());
		int tb=sz(b);
		vector<int> acum(tb);
		acum[0]=b[0];
		forr(i,1,tb) acum[i]=acum[i-1]+b[i];
		//~ forn(i,tb) cout << b[i] << " ";
		//~ cout << endl;
		int cant=0;
		int suma=0;
		int ans=n;
		dforn(i,tb){
			int auxacum=0;
			if(i) auxacum=acum[i-1];
			//~ cout << auxacum << " " << suma-b[i]*cant << endl;
			int pans = auxacum+(suma-b[i]*cant);
			cant++;
			suma+=b[i];
			ans=min(pans,ans);
		}
		cout << ans << '\n';
	}
	
	#ifdef ANARAP
	t1 = clock();double time = (double(t1-t0)/CLOCKS_PER_SEC); cout << "Execution Time: " << time << endl;
	#endif
	return 0;
}

```