#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main ()
{
  int T;
  cin>>T;
  while(T--){
    ll n;
    cin>>n;
    vector<ll> a(n);
    ll sum = n;
    for(ll i=0;i<n;i++) {cin >> a[i];}
    ll ans = n;
    vector<pair<ll,ll>> v;
    map<ll,ll> m;
    for(auto x : a) m[x]++;
    for(auto it : m) v.push_back({it.second,it.first});
    sort(v.begin(),v.end());
    ll t=v.size();
    ll p = -1;
    for(ll k=v.size()-1;k>=0;k--){
      if(ans > n-v[k].first*(t-k)){
        ans = n-v[k].first*(t-k);
        p = k;
      }
    }
    cout << ans << endl;
  }
  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>
using namespace std;
typedef long long ll;
int main ()
{
  int T;
  cin>>T;
  while(T--){
    ll n;
    cin>>n;
    vector<ll> a(n);
    ll sum = n;
    for(ll i=0;i<n;i++) {cin >> a[i];}
    ll ans = n;
    vector<pair<ll,ll>> v;
    map<ll,ll> m;
    for(auto x : a) m[x]++;
    for(auto it : m) v.push_back({it.second,it.first});
    sort(v.begin(),v.end());
    ll t=v.size();
    ll p = -1;
    for(ll k=v.size()-1;k>=0;k--){
      if(ans > n-v[k].first*(t-k)){
        ans = n-v[k].first*(t-k);
        p = k;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```