#include<bits/stdc++.h>
#define int long long
#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define f(i,n) for(int i=0;i<n;i++)
#define pb push_back
#define mp make_pair
#define endl '\n'

using namespace std;
void SieveOfEratosthenes(int n) 
{ 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
    for (int p=2; p<=n; p++) 
       if (prime[p]) 
          cout << p << " "; 
} 
bool isPow(int n)
{
   if(n==0)
   return false;
   return (ceil(log2(n))==floor(log2(n)));
}
void solve(int a[],int l,int r,int cnt,int n,int d[])
{
    if(l>r) return;
    int max=-1,ind;
    //cout<<l<<" "<<r<<endl;
    for(int i=l;i<=r;i++)
    {
        if(a[i]>max)
        {
            max=a[i];ind=i;
        }
    }
    //cout<<max<<" "<<ind<<" "<<cnt<<endl;
    d[max]=cnt;
    solve(a,l,ind-1,cnt+1,n,d);
    solve(a,ind+1,r,cnt+1,n,d);
}
signed main(){
    optimize();
    int t;
    cin >> t;
    while(t--)
    {
        vector <int> v;
        int n,sum=0;cin>>n;
        int a[n],b[n],m=0;
        map <int,int> mp;
        for(int i=0;i<n;i++) 
        {
            cin>>a[i];
            mp[a[i]]++;
        }
        for(auto i=mp.begin();i!=mp.end();i++)
        {
            b[m]=i->second;m++;
        }
        sort(b,b+m);
        int c[m],d[m];
        for(int i=0;i<m;i++)
        {
            sum+=b[i];
            c[i]=sum;
        }
        sum=0;
        for(int i=m-1;i>=0;i--)
        {
            sum+=b[i];
            d[i]=sum;
        }
        int min=n,x=0;
        for(int i=0;i<m-1;i++)
        {
            int y=x+d[i+1]-(b[i]*(m-i-1));
            if(y<min) min=y;
            if(b[i]!=b[i+1])
            x=c[i];
        }
        if(m!=1&&c[m-2]<min) min=c[m-2];
        if(m==1) min=0;
        cout<<min<<endl;
    }
    return 0;
}

### Prompt

Develop a solution in CPP 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>
#define int long long
#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define f(i,n) for(int i=0;i<n;i++)
#define pb push_back
#define mp make_pair
#define endl '\n'

using namespace std;
void SieveOfEratosthenes(int n) 
{ 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
    for (int p=2; p<=n; p++) 
       if (prime[p]) 
          cout << p << " "; 
} 
bool isPow(int n)
{
   if(n==0)
   return false;
   return (ceil(log2(n))==floor(log2(n)));
}
void solve(int a[],int l,int r,int cnt,int n,int d[])
{
    if(l>r) return;
    int max=-1,ind;
    //cout<<l<<" "<<r<<endl;
    for(int i=l;i<=r;i++)
    {
        if(a[i]>max)
        {
            max=a[i];ind=i;
        }
    }
    //cout<<max<<" "<<ind<<" "<<cnt<<endl;
    d[max]=cnt;
    solve(a,l,ind-1,cnt+1,n,d);
    solve(a,ind+1,r,cnt+1,n,d);
}
signed main(){
    optimize();
    int t;
    cin >> t;
    while(t--)
    {
        vector <int> v;
        int n,sum=0;cin>>n;
        int a[n],b[n],m=0;
        map <int,int> mp;
        for(int i=0;i<n;i++) 
        {
            cin>>a[i];
            mp[a[i]]++;
        }
        for(auto i=mp.begin();i!=mp.end();i++)
        {
            b[m]=i->second;m++;
        }
        sort(b,b+m);
        int c[m],d[m];
        for(int i=0;i<m;i++)
        {
            sum+=b[i];
            c[i]=sum;
        }
        sum=0;
        for(int i=m-1;i>=0;i--)
        {
            sum+=b[i];
            d[i]=sum;
        }
        int min=n,x=0;
        for(int i=0;i<m-1;i++)
        {
            int y=x+d[i+1]-(b[i]*(m-i-1));
            if(y<min) min=y;
            if(b[i]!=b[i+1])
            x=c[i];
        }
        if(m!=1&&c[m-2]<min) min=c[m-2];
        if(m==1) min=0;
        cout<<min<<endl;
    }
    return 0;
}

```