#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef vector<vector<ll>> vvl;
typedef vector<ll> vll;
#define F(i,a,n) for(ll i=a;i<n;i++)
#define mod 1000000007
#define mod2 998244353

#define INF (1e18)+5
bool sq(ll c){
	ll z=sqrt(c);
	if(z*z==c)return true;
	else return false;
}
bool cube_root(ll x){
	ll l=1ll,r=10000;
	while(l<=r){
		ll m=(l+r)/2;
		if(m*m*m==x){
			return true;
		}
		else if(m*m*m>x){
			r=m-1;
		}
		else {
			l=m+1;
		}
	}
	return false;
}
void solve(){
	ll n;
	cin>>n;
	vll v(n);
	map<ll,ll>m,m2;
	F(i,0,n){cin>>v[i];
	m[v[i]]++;
	}
	for(auto i:m){
		m2[i.second]++;
	}
	ll ans=0,s=0,y=0;
	vector<pll>x;
	for(auto i:m2){x.push_back(i);}
	sort(x.begin(),x.end(),greater<pll>());
	// for(auto i:x){cout<<i.first<<" "<<i.second<<endl;}
	for(auto i:x){
		ans=max(ans,s*i.first+i.second*i.first);
		s+=i.second;
		// y+=i.first;
		// cout<<ans<<" "<<i.second<<" ";
	}
	// ans=min(ans,n-s);

	cout<<n-ans<<endl;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	#ifndef ONLINE_JUDGE
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
	#endif
	ll w=1;
	cin>>w;
	F(i,0,w){
		solve();
	}
	return 0;
}
### Prompt

Please formulate 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>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef vector<vector<ll>> vvl;
typedef vector<ll> vll;
#define F(i,a,n) for(ll i=a;i<n;i++)
#define mod 1000000007
#define mod2 998244353

#define INF (1e18)+5
bool sq(ll c){
	ll z=sqrt(c);
	if(z*z==c)return true;
	else return false;
}
bool cube_root(ll x){
	ll l=1ll,r=10000;
	while(l<=r){
		ll m=(l+r)/2;
		if(m*m*m==x){
			return true;
		}
		else if(m*m*m>x){
			r=m-1;
		}
		else {
			l=m+1;
		}
	}
	return false;
}
void solve(){
	ll n;
	cin>>n;
	vll v(n);
	map<ll,ll>m,m2;
	F(i,0,n){cin>>v[i];
	m[v[i]]++;
	}
	for(auto i:m){
		m2[i.second]++;
	}
	ll ans=0,s=0,y=0;
	vector<pll>x;
	for(auto i:m2){x.push_back(i);}
	sort(x.begin(),x.end(),greater<pll>());
	// for(auto i:x){cout<<i.first<<" "<<i.second<<endl;}
	for(auto i:x){
		ans=max(ans,s*i.first+i.second*i.first);
		s+=i.second;
		// y+=i.first;
		// cout<<ans<<" "<<i.second<<" ";
	}
	// ans=min(ans,n-s);

	cout<<n-ans<<endl;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	#ifndef ONLINE_JUDGE
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
	#endif
	ll w=1;
	cin>>w;
	F(i,0,w){
		solve();
	}
	return 0;
}
```