Problem: This problem is actually a subproblem of problem G from the same contest.

There are n candies in a candy box. The type of the i-th candy is a_i (1 ≤ a_i ≤ n).

You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad). 

It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.

Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.

You have to answer q independent queries.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries. Each query is represented by two lines.

The first line of each query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of candies.

The second line of each query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the type of the i-th candy in the box.

It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.

Output

For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.

Example

Input


3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7


Output


3
10
9

Note

In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.

Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[200001];
int main() {
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    int x;
    cin >> x;
    for (int c = 0; c < x; c++) cin >> a[c];
    map<int, int> Q;
    for (int j = 0; j < x; j++) Q[a[j]]++;
    map<int, int>::iterator iter;
    map<int, int> W;
    for (iter = Q.begin(); iter != Q.end(); iter++) {
      W[iter->second]++;
    }
    map<int, int>::reverse_iterator zz;
    zz = W.rbegin();
    int f = zz->first;
    int s = zz->second;
    long long sum = 0;
    for (; zz != W.rend();) {
      zz++;
      if (zz == W.rend()) {
        long long hh = min(f, s);
        sum = sum + (f + f - hh + 1) * hh / 2;
      } else if (f - (zz->first) <= s) {
        sum = sum + (f + zz->first + 1) * (f - (zz->first)) / 2;
        zz->second = zz->second + s - (f - (zz->first));
      } else
        sum = sum + (f * 2 - s + 1) * s / 2;
      f = zz->first;
      s = zz->second;
    }
    cout << sum << endl;
  }
  return 0;
}