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 hcf(long long a, long long b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
bool compare(pair<int, int> p, pair<int, int> q) {
  if (p.first == q.first)
    return p.second > q.second;
  else
    return p.first > q.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<int, int>> count(n + 1, {0, 0});
    for (int i = 0; i < n; i++) {
      int index, keep;
      cin >> index >> keep;
      count[index].first++;
      count[index].second += keep;
    }
    sort(count.begin(), count.end(), compare);
    long long ans = 0;
    long long kitne = 0;
    int till = n;
    for (int i = 0; i < n; i++) {
      if (count[i].first == 0) break;
      sort(count.begin() + i, count.begin() + till, compare);
      ans += count[i].first;
      kitne += min(count[i].first, count[i].second);
      int j = i + 1;
      while (j < n && count[j].first == count[i].first) {
        count[j].first--;
        j++;
      }
      while (j < n && count[j].first + 1 == count[i].first) j++;
      till = j;
    }
    cout << ans << " " << kitne << endl;
  }
  return 0;
}