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;
const int N = 200001;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> f(n + 1);
  for (int i = 0; i < n; i++) {
    int c;
    cin >> c;
    f[c].first++;
    int c2;
    cin >> c2;
    if (c2 == 1) f[c].second++;
  }
  sort(f.begin(), f.end(), [&](pair<int, int> p1, pair<int, int> p2) {
    if (p2.first == p1.first) return p2.second < p1.second;
    return p2.first < p1.first;
  });
  int ans = 0;
  int ans2 = 0;
  int last = f[0].first;
  for (int j = 0; j < n; j++) {
    if (f[j].first >= last) {
      ans += last;
      int idx = j;
      while (idx < n && f[idx].first >= last) idx++;
      sort(f.begin() + j, f.begin() + idx,
           [&](pair<int, int> p1, pair<int, int> p2) {
             return p2.second < p1.second;
           });
      ans2 += min(last, f[j].second);
      last--;
    } else {
      ans += f[j].first;
      ans2 += f[j].second;
      last = f[j].first - 1;
    }
    if (f[j].first == 0 || last <= 0) break;
  }
  cout << ans << " " << ans2 << endl;
}
int main() {
  cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int q;
  cin >> q;
  for (int t = 0; t < q; t++) {
    solve();
  }
}