#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename... Args> void read(Args&... args) { ((cin >> args), ...); }
template<typename... Args> void write(Args... args) { ((cout << args << " "), ...); }
template<typename... Args> void writeln(Args... args) { write(args...); cout << '\n'; }
#define all(x) (x).begin(), (x).end()
#define pb push_back

void solve_case(){
    int n; cin >> n;
    vector<int> a(n);
    map<int,int> oc;
    for(int i = 0; i<n; i++) {
        int v; cin >> v;
        oc[v]++;
    }
    vector<int> freq;
    for(auto i : oc) freq.push_back(i.second);
    sort(all(freq));
    vector<int> pre(freq.size());
    pre[0] = freq[0];
    for(int i = 1; i<freq.size(); i++) pre[i] = pre[i-1] + freq[i];

    int ans = 1000000;
    for(int i = 0; i<freq.size(); i++){
        int amountAfter = freq.size()-i-1;
        int afterSum = pre[freq.size()-1]-pre[i];
        int removeAfter = max(0,afterSum-amountAfter*freq[i]);
        int l = 0, r = i, last = -1;
        while(l<=r){
            int md = (l+r)/2;
            if(freq[md]<freq[i]){
                last = md;
                l = md+1;
            }else r = md-1;
        }
        int removeBefore = 0;
        if(last!=-1) removeBefore = pre[last];
        ans = min(ans,removeBefore+removeAfter);
    }
    cout << ans << endl;
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t;
    cin >> t;
    while(t-->0){
        solve_case();
    }
}
### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename... Args> void read(Args&... args) { ((cin >> args), ...); }
template<typename... Args> void write(Args... args) { ((cout << args << " "), ...); }
template<typename... Args> void writeln(Args... args) { write(args...); cout << '\n'; }
#define all(x) (x).begin(), (x).end()
#define pb push_back

void solve_case(){
    int n; cin >> n;
    vector<int> a(n);
    map<int,int> oc;
    for(int i = 0; i<n; i++) {
        int v; cin >> v;
        oc[v]++;
    }
    vector<int> freq;
    for(auto i : oc) freq.push_back(i.second);
    sort(all(freq));
    vector<int> pre(freq.size());
    pre[0] = freq[0];
    for(int i = 1; i<freq.size(); i++) pre[i] = pre[i-1] + freq[i];

    int ans = 1000000;
    for(int i = 0; i<freq.size(); i++){
        int amountAfter = freq.size()-i-1;
        int afterSum = pre[freq.size()-1]-pre[i];
        int removeAfter = max(0,afterSum-amountAfter*freq[i]);
        int l = 0, r = i, last = -1;
        while(l<=r){
            int md = (l+r)/2;
            if(freq[md]<freq[i]){
                last = md;
                l = md+1;
            }else r = md-1;
        }
        int removeBefore = 0;
        if(last!=-1) removeBefore = pre[last];
        ans = min(ans,removeBefore+removeAfter);
    }
    cout << ans << endl;
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t;
    cin >> t;
    while(t-->0){
        solve_case();
    }
}
```