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;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vii = vector<vi>;
int sum(int n) {
  int s = 0;
  while (n) {
    s += n % 10;
    n /= 10;
  }
  return s;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    vector<int> kek(n + 1);
    vector<int> counter(n + 1);
    for (int i = 0; i < n; ++i) {
      int v;
      cin >> v;
      --counter[kek[v]];
      ++kek[v];
      ++counter[kek[v]];
    }
    ll ans = 0;
    vector<int> unused;
    for (int i = 1; i <= n; ++i) {
      if (counter[i]) {
        ans += i;
        for (int j = 1; j < counter[i]; ++j) {
          if (unused.empty()) break;
          ans += unused[unused.size() - 1];
          unused.erase(unused.end() - 1);
        }
      } else
        unused.push_back(i);
    }
    cout << ans << '\n';
  }
  return 0;
}