/*
Created by : Ivan Deka
About me : Just a coder trying to code out the best of me!
*/
#include<bits/stdc++.h>
using namespace std;

typedef int long long ll;
typedef unsigned int long long ul;
typedef long double ld;

#define mod 1000000007
#define mod1 998244353
#define pb push_back
#define mk make_pair
#define vi vector<ll>
#define pii pair<ll, ll>
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define sz(a) a.size()
#define ub(a,x) upper_bound(all(a),x) - a.begin()
#define lb(a,x) lower_bound(all(a),x) - a.begin()

ll inline powerM(ll a, ll b, ll p){
    a %= p;
    ll ans = 1;
    while(b>0){
        if(b & 1)
            ans = (ans*a)%p;
        a = (a*a)%p;
        b >>= 1;
    }
    return ans;
}

ll inv(ll n, ll p){
    return powerM(n,p-2, p);
}

ll inline power(ll a, ll b){
    //a %= p;
    ll ans = 1;
    while(b>0){
        if(b & 1)
            ans = (ans*a);
        a = (a*a);
        b >>= 1;
    }
    return ans;
}

void solve()
{
    ll n;
    cin>>n;
    ll a[n];
    map<ll,ll> mp,mp2,mp1;
    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        mp[a[i]]++;
    }

    for(ll i=0;i<n;i++)
    {
        if(mp1[a[i]]==0) mp2[mp[a[i]]]++;
        mp1[a[i]]++;
    }
    ll mn=0;
    ll val=0;
    for(auto it=mp2.rbegin();it!=mp2.rend();it++)
    {
        ll w=it->ff;
        val+=it->ss;
        mn=max(mn,w*(val));
    }
    cout<<n-mn<<"\n";
}

int main()
{
    //ϳᕱᎥ ♅ᕱภuოᕱภ
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    ll t;
    cin>>t;
    for(ll i=1;i<=t;i++)
    {
        //cout<<"Case #"<<i<<":"<<" ";
        solve();
    }
}
### Prompt

Construct a CPP code solution to the problem outlined:

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
/*
Created by : Ivan Deka
About me : Just a coder trying to code out the best of me!
*/
#include<bits/stdc++.h>
using namespace std;

typedef int long long ll;
typedef unsigned int long long ul;
typedef long double ld;

#define mod 1000000007
#define mod1 998244353
#define pb push_back
#define mk make_pair
#define vi vector<ll>
#define pii pair<ll, ll>
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define sz(a) a.size()
#define ub(a,x) upper_bound(all(a),x) - a.begin()
#define lb(a,x) lower_bound(all(a),x) - a.begin()

ll inline powerM(ll a, ll b, ll p){
    a %= p;
    ll ans = 1;
    while(b>0){
        if(b & 1)
            ans = (ans*a)%p;
        a = (a*a)%p;
        b >>= 1;
    }
    return ans;
}

ll inv(ll n, ll p){
    return powerM(n,p-2, p);
}

ll inline power(ll a, ll b){
    //a %= p;
    ll ans = 1;
    while(b>0){
        if(b & 1)
            ans = (ans*a);
        a = (a*a);
        b >>= 1;
    }
    return ans;
}

void solve()
{
    ll n;
    cin>>n;
    ll a[n];
    map<ll,ll> mp,mp2,mp1;
    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        mp[a[i]]++;
    }

    for(ll i=0;i<n;i++)
    {
        if(mp1[a[i]]==0) mp2[mp[a[i]]]++;
        mp1[a[i]]++;
    }
    ll mn=0;
    ll val=0;
    for(auto it=mp2.rbegin();it!=mp2.rend();it++)
    {
        ll w=it->ff;
        val+=it->ss;
        mn=max(mn,w*(val));
    }
    cout<<n-mn<<"\n";
}

int main()
{
    //ϳᕱᎥ ♅ᕱภuოᕱภ
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    ll t;
    cin>>t;
    for(ll i=1;i<=t;i++)
    {
        //cout<<"Case #"<<i<<":"<<" ";
        solve();
    }
}
```