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 main() {
  long long int q;
  cin >> q;
  for (long long int t = 0; t < q; ++t) {
    long long int n;
    cin >> n;
    long long int a[n];
    set<long long int> s;
    map<long long int, long long int> m;
    for (long long int i = 0; i < n; ++i) {
      cin >> a[i];
      s.insert(a[i]);
      if (m.find(a[i]) != m.end()) {
        ++m[a[i]];
      } else {
        m.insert(pair<long long int, long long int>(a[i], 1));
      }
    }
    vector<long long int> v;
    auto r = s.begin();
    for (r = s.begin(); r != s.end(); ++r) {
      v.push_back(m[*r]);
    }
    sort(v.begin(), v.end());
    long long int zx = v.size();
    long long int max = v[zx - 1], klj = 0;
    long long int sum = v[zx - 1];
    for (long long int i = zx - 2; i >= 0 && klj == 0; --i) {
      if (v[i] >= max) {
        sum += (max - 1);
        --max;
      } else {
        sum += v[i];
        max = v[i];
      }
      if (max <= 0) ++klj;
    }
    cout << sum << endl;
  }
}