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() {
  int n, i, j, t;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    int a, f;
    pair<int, int> has[n + 1];
    for (i = 1; i < n + 1; i++) has[i].first = has[i].second = 0;
    for (i = 0; i < n; i++) {
      cin >> a >> f;
      has[a].first++;
      has[a].second += f;
    }
    vector<pair<int, int> > v;
    for (i = 1; i < n + 1; i++) {
      if (has[i].first) v.push_back(has[i]);
    }
    sort(v.begin(), v.end());
    vector<int> take;
    take.push_back(v.back().first);
    int ans = v.back().first;
    int last = v.back().first;
    for (i = v.size() - 2; i >= 0; i--) {
      if (last == 1) break;
      if (v[i].first < last) {
        ans += v[i].first;
        last = v[i].first;
        take.push_back(v[i].first);
      } else {
        last--;
        ans += last;
        take.push_back(last);
      }
    }
    priority_queue<pair<int, int> > pq;
    int l = v.size() - 1;
    pair<int, int> z;
    int ans2 = 0;
    for (i = 0; i < take.size(); i++) {
      while (l >= 0 && v[l].first >= take[i]) {
        pq.push({v[l].second, v[l].first});
        l--;
      }
      z = pq.top();
      ans2 += min(z.first, take[i]);
      pq.pop();
    }
    cout << ans << ' ' << ans2 << '\n';
  }
  return 0;
}