#include <bits/stdc++.h>
#include <iostream>
#include <math.h>
#include <vector>
#include <deque>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stdio.h>
#include <cstdlib>
#include <random>
#include <chrono>
#include <fstream>
using namespace std;
typedef unsigned long long int ul;
typedef long long int ll;
typedef long double ld;
mt19937 rng((unsigned int)chrono::steady_clock::now().time_since_epoch().count());
#define mod 1000000007LL
#define sgraph 100100LL
#define segment 800100LL
#define all(V) V.begin(),V.end()
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll T;
    cin>>T;
    while(T--){
        ll n;
        cin>>n;
        vector<ll> a(n);
        set<ll> st;
        map<ll,ll> mp;
        for(ll i=0;i<n;i++){
            cin>>a[i];
            st.insert(a[i]);
        }
        ll itr=0;
        while(st.size()){
            mp[*st.begin()]=itr++;
            st.erase(st.begin());
        }
        vector<ll> hash(itr);
        for(ll i=0;i<n;i++){
            a[i]=mp[a[i]];
            hash[a[i]]++;
        }
        ll ans=n;
        sort(all(hash));
        ll prev=0,cur=0;
        for(ll i=0;i<itr;i++){
            ll j=i;
            while(j<itr&&hash[i]==hash[j]){
                j++;
            }
            ans=min(ans,n-((itr-i)*hash[i]));
            i=j-1;
        }
        cout<<ans<<"\n";
        
    }
}

### 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>
#include <iostream>
#include <math.h>
#include <vector>
#include <deque>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stdio.h>
#include <cstdlib>
#include <random>
#include <chrono>
#include <fstream>
using namespace std;
typedef unsigned long long int ul;
typedef long long int ll;
typedef long double ld;
mt19937 rng((unsigned int)chrono::steady_clock::now().time_since_epoch().count());
#define mod 1000000007LL
#define sgraph 100100LL
#define segment 800100LL
#define all(V) V.begin(),V.end()
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll T;
    cin>>T;
    while(T--){
        ll n;
        cin>>n;
        vector<ll> a(n);
        set<ll> st;
        map<ll,ll> mp;
        for(ll i=0;i<n;i++){
            cin>>a[i];
            st.insert(a[i]);
        }
        ll itr=0;
        while(st.size()){
            mp[*st.begin()]=itr++;
            st.erase(st.begin());
        }
        vector<ll> hash(itr);
        for(ll i=0;i<n;i++){
            a[i]=mp[a[i]];
            hash[a[i]]++;
        }
        ll ans=n;
        sort(all(hash));
        ll prev=0,cur=0;
        for(ll i=0;i<itr;i++){
            ll j=i;
            while(j<itr&&hash[i]==hash[j]){
                j++;
            }
            ans=min(ans,n-((itr-i)*hash[i]));
            i=j-1;
        }
        cout<<ans<<"\n";
        
    }
}

```