#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()



void solve(){
    
    int n; cin >> n;
    int a[n];
    map< int, int > cnt2, freq, freq2;
    for(int& u : a){
        cin >> u;
        ++cnt2[u];
    }
    for(auto u : cnt2){
        ++freq[u.ss];
        freq2[u.ss] += u.ss;
    }
    //dp[i] = number of elements that I have to remove from higher frequencies to make them
    //equal to frequency i
    //cnt[i] = number of distinct elements with freq >= i
    int cnt[n + 2]; cnt[n + 1] = 0;
    for(int i = n; i >= 1; --i){
        cnt[i] = cnt[i + 1] + freq[i];
    }
    int dp[n + 2]; dp[n + 1] = 0;
    for(int i = n; i >= 1; --i){
        dp[i] = dp[i + 1] + cnt[i + 1];
    }
    int ans = n;
    int lessThan = 0;
    for(int i = 1; i <= n; ++i){
        ans = min(ans, lessThan + dp[i]);
        lessThan += freq2[i];
    }
    cout << ans << '\n';

}


int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false); cin.tie(0);

    int t = 1;
    cin >> t;


    for(int i = 1; i <= t; i++){
        // cout << "Case #" << i << ": ";

        solve();

    }
    return 0;
}
### Prompt

In Cpp, your task is to solve 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;

#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()



void solve(){
    
    int n; cin >> n;
    int a[n];
    map< int, int > cnt2, freq, freq2;
    for(int& u : a){
        cin >> u;
        ++cnt2[u];
    }
    for(auto u : cnt2){
        ++freq[u.ss];
        freq2[u.ss] += u.ss;
    }
    //dp[i] = number of elements that I have to remove from higher frequencies to make them
    //equal to frequency i
    //cnt[i] = number of distinct elements with freq >= i
    int cnt[n + 2]; cnt[n + 1] = 0;
    for(int i = n; i >= 1; --i){
        cnt[i] = cnt[i + 1] + freq[i];
    }
    int dp[n + 2]; dp[n + 1] = 0;
    for(int i = n; i >= 1; --i){
        dp[i] = dp[i + 1] + cnt[i + 1];
    }
    int ans = n;
    int lessThan = 0;
    for(int i = 1; i <= n; ++i){
        ans = min(ans, lessThan + dp[i]);
        lessThan += freq2[i];
    }
    cout << ans << '\n';

}


int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(false); cin.tie(0);

    int t = 1;
    cin >> t;


    for(int i = 1; i <= t; i++){
        // cout << "Case #" << i << ": ";

        solve();

    }
    return 0;
}
```