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

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
#define FOR(i, n) for(int (i)=0; (i)<(n); (i)++)
#define FOR1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FORI(i, n) for(int (i)=n-1; (i)>=0; (i)--)
template<class T, class U> void umin(T& x, const U& y){ x = min(x, (T)y);}
template<class T, class U> void umax(T& x, const U& y){ x = max(x, (T)y);}
template<class T, class U> void init(vector<T> &v, U x, size_t n) { v=vector<T>(n, (T)x); }
template<class T, class U, typename... W> void init(vector<T> &v, U x, size_t n, W... m) { v=vector<T>(n);  for(auto& a : v) init(a, x, m...); }


int main(int argc, char** argv) {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cout << setprecision(15);
    if (argc == 2 && atoi(argv[1]) == 123456789) freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);
    int T = 1;
    cin >> T;
    FOR1(t, T){
        int n;
        cin >> n;
        vi a(n);
        map<int, int> H;
        FOR(i, n) {
            cin >> a[i];
            H[a[i]]++;
        }
        map<int , int> H2; //cnt of cnts
        int groups = H.size();
        for(auto it : H) {
            H2[it.second]++;
        }

//        int removed = 0;
        int sol = n;
        for(auto it : H2){
            umin(sol, n - groups * it.first);
            groups -= it.second;
//            removed += it.first * it.second;
        }
        cout << sol << endl;
    }


    if (argc == 2 && atoi(argv[1]) == 123456789) cout << clock()*1.0/CLOCKS_PER_SEC << " sec\n";
    return 0;
}


### Prompt

Please provide a Cpp coded solution 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;
typedef pair<int, int> pii;
typedef vector<int> vi;
#define FOR(i, n) for(int (i)=0; (i)<(n); (i)++)
#define FOR1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FORI(i, n) for(int (i)=n-1; (i)>=0; (i)--)
template<class T, class U> void umin(T& x, const U& y){ x = min(x, (T)y);}
template<class T, class U> void umax(T& x, const U& y){ x = max(x, (T)y);}
template<class T, class U> void init(vector<T> &v, U x, size_t n) { v=vector<T>(n, (T)x); }
template<class T, class U, typename... W> void init(vector<T> &v, U x, size_t n, W... m) { v=vector<T>(n);  for(auto& a : v) init(a, x, m...); }


int main(int argc, char** argv) {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cout << setprecision(15);
    if (argc == 2 && atoi(argv[1]) == 123456789) freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);
    int T = 1;
    cin >> T;
    FOR1(t, T){
        int n;
        cin >> n;
        vi a(n);
        map<int, int> H;
        FOR(i, n) {
            cin >> a[i];
            H[a[i]]++;
        }
        map<int , int> H2; //cnt of cnts
        int groups = H.size();
        for(auto it : H) {
            H2[it.second]++;
        }

//        int removed = 0;
        int sol = n;
        for(auto it : H2){
            umin(sol, n - groups * it.first);
            groups -= it.second;
//            removed += it.first * it.second;
        }
        cout << sol << endl;
    }


    if (argc == 2 && atoi(argv[1]) == 123456789) cout << clock()*1.0/CLOCKS_PER_SEC << " sec\n";
    return 0;
}


```