#include <bits/stdc++.h>

#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define sz(x) (int)x.size()




using namespace std;


using ll  = long long;
using pii = pair<int,int>;
using ull = unsigned long long;

void err(istream_iterator<string> it) {cerr << endl;}
template<typename T, typename... Args>void err(istream_iterator<string> it, T a, Args... args) {cerr << *it << " = " << a << endl;err(++it, args...);}




const int dx[] = {0, -1, 1,  0, 1, -1,  1, -1};
const int dy[] = {1 , 0, 0, -1, 1, -1, -1,  1};

const double EPS = 1e-9;
const int mod = 1e9 + 7;
const long long INF = 1e18;
const int N = 2e5 + 5;


int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int q;
    cin >> q;
    while(q--){
            int n;
            cin >> n;
            int a;
            map<int,int>frq1, frq2;
            for(int i = 1; i <= n; ++i){
                    cin >> a;
                    frq1[a]++;
            }
            for(auto &i : frq1){
                frq2[i.second]++;
            }
            ll ans = n-1;
            vector<pair<ll,ll> > v;
            v.push_back({0,0});
            for(auto i : frq2){
                v.push_back({i.first * i.second, i.second});
            }
            for(int i = 1; i < v.size(); ++i){
                    v[i].first += v[i - 1].first;
                    v[i].second  += v[i - 1].second;
            }
            ll cnt = 0;
            for(int i = 1; i < v.size(); ++i){
                ll c = v[sz(v)-1].first-v[i].first;
                ll toR = v[sz(v)-1].second - v[i].second;
                ll cur = (v[i].first - v[i - 1].first) / (v[i].second - v[i - 1].second);
                ans = min(ans , cnt + (c - cur * toR));
                cnt= v[i].first;
            }

            cout << ans << '\n';
    }
    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>

#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define sz(x) (int)x.size()




using namespace std;


using ll  = long long;
using pii = pair<int,int>;
using ull = unsigned long long;

void err(istream_iterator<string> it) {cerr << endl;}
template<typename T, typename... Args>void err(istream_iterator<string> it, T a, Args... args) {cerr << *it << " = " << a << endl;err(++it, args...);}




const int dx[] = {0, -1, 1,  0, 1, -1,  1, -1};
const int dy[] = {1 , 0, 0, -1, 1, -1, -1,  1};

const double EPS = 1e-9;
const int mod = 1e9 + 7;
const long long INF = 1e18;
const int N = 2e5 + 5;


int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int q;
    cin >> q;
    while(q--){
            int n;
            cin >> n;
            int a;
            map<int,int>frq1, frq2;
            for(int i = 1; i <= n; ++i){
                    cin >> a;
                    frq1[a]++;
            }
            for(auto &i : frq1){
                frq2[i.second]++;
            }
            ll ans = n-1;
            vector<pair<ll,ll> > v;
            v.push_back({0,0});
            for(auto i : frq2){
                v.push_back({i.first * i.second, i.second});
            }
            for(int i = 1; i < v.size(); ++i){
                    v[i].first += v[i - 1].first;
                    v[i].second  += v[i - 1].second;
            }
            ll cnt = 0;
            for(int i = 1; i < v.size(); ++i){
                ll c = v[sz(v)-1].first-v[i].first;
                ll toR = v[sz(v)-1].second - v[i].second;
                ll cur = (v[i].first - v[i - 1].first) / (v[i].second - v[i - 1].second);
                ans = min(ans , cnt + (c - cur * toR));
                cnt= v[i].first;
            }

            cout << ans << '\n';
    }
    return 0;
}

```