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;
long long mx;
bool cmp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (min(a.first, mx - 1) != min(b.first, mx - 1))
    return min(a.first, mx - 1) < min(b.first, mx - 1);
  else
    return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t, n, m, ans, f, candy, b;
  cin >> t;
  while (t--) {
    cin >> n;
    mx = 0;
    vector<pair<long long, long long>> data(n + 1);
    for (int i = 0; i < n; ++i) {
      cin >> candy >> b;
      ++data[candy].first;
      data[candy].second += b;
      mx = max(mx, data[candy].first);
    }
    sort(data.begin(), data.end());
    priority_queue<pair<long long, long long>> q;
    ans = f = 0;
    ++mx;
    while (!data.empty() && mx > 0) {
      while (!q.empty() && (data.back().first != mx - 1)) {
        b = min(q.top().second, mx - 1);
        ans += b;
        f += min(b, q.top().first);
        mx = b;
        q.pop();
      }
      mx = data.back().first + 1;
      while (!data.empty() && (data.back().first == (mx - 1))) {
        q.push({data.back().second, data.back().first});
        data.pop_back();
      }
    }
    cout << ans << " " << f << "\n";
  }
  return 0;
}