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
#include <iostream>
#include <algorithm>

using namespace std;

#ifndef ONLINE_JUDGE
  #include <fstream>
  #define cin fin
  #define cout fout
  ifstream fin ("test.in");
  ofstream fout ("test.out");
#endif

int t, n, a[200001], f[200001], nra, p[100001], k;

int rasp (int mij) {
  int r = 0;
  for (int i = 1; i <= k; i++) {
    if (p[i] - mij < 0)
      r += p[i] * f[p[i]];
    else
      r += (p[i] - mij) * f[p[i]];
  }
  return r;
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(NULL);

  cin >> t;
  for (int tt = 1; tt <= t; tt++) {
    cin >> n; int r = n + 1; k = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i]; f[i] = 0;
    }
    sort (a + 1, a + n + 1);
    nra = 0; a[0] = a[1]; a[n + 1] = a[n] + 1;
    for (int i = 1; i <= n + 1; i++) {
      if (a[i] == a[i - 1])
        nra++;
      else {
        f[nra]++;
        nra = 1;
      }
    }
    for (int i = 1; i <= n; i++)
      if (f[i])
        p[++k] = i;
    for (int i = 1; i <= k; i++)
      r = min (r, rasp(p[i]));
    cout << r << '\n';
  }
  return 0;
}
