#include <bits/stdc++.h>
using namespace std;
map<long long, pair<long long, long long> > Map;
vector<pair<long long, long long> > v;
long long a[200000];
bool used[200000];
bool f[200000];
set<pair<long long, long long> > Set;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 0;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long n = 0;
    cin >> n;
    Map.clear();
    for (long long i = 0; i < n; i++) {
      cin >> a[i] >> f[i];
      Map[a[i]].first++;
      Map[a[i]].second += f[i];
    }
    v.clear();
    for (auto p : Map) {
      v.push_back(p.second);
      used[v.size() - 1] = false;
    }
    sort(v.rbegin(), v.rend());
    long long cur_c = INT_MAX, res = 0, res2 = 0, lst = -1;
    for (long long i = 0; i < v.size(); i++) {
      if (v[i].first < cur_c)
        cur_c = v[i].first;
      else
        cur_c = max(0ll, cur_c - 1);
      long long ind = -1, Max = 0;
      for (long long j = lst + 1; j < v.size(); j++) {
        if (v[j].first < cur_c) break;
        if (v[j].first >= cur_c) {
          Set.insert({-v[j].second, j});
        }
        lst = j;
      }
      res += cur_c;
      res2 += min(cur_c, -Set.begin()->first);
      Set.erase(Set.begin());
    }
    cout << res << ' ' << res2 << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
map<long long, pair<long long, long long> > Map;
vector<pair<long long, long long> > v;
long long a[200000];
bool used[200000];
bool f[200000];
set<pair<long long, long long> > Set;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 0;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long n = 0;
    cin >> n;
    Map.clear();
    for (long long i = 0; i < n; i++) {
      cin >> a[i] >> f[i];
      Map[a[i]].first++;
      Map[a[i]].second += f[i];
    }
    v.clear();
    for (auto p : Map) {
      v.push_back(p.second);
      used[v.size() - 1] = false;
    }
    sort(v.rbegin(), v.rend());
    long long cur_c = INT_MAX, res = 0, res2 = 0, lst = -1;
    for (long long i = 0; i < v.size(); i++) {
      if (v[i].first < cur_c)
        cur_c = v[i].first;
      else
        cur_c = max(0ll, cur_c - 1);
      long long ind = -1, Max = 0;
      for (long long j = lst + 1; j < v.size(); j++) {
        if (v[j].first < cur_c) break;
        if (v[j].first >= cur_c) {
          Set.insert({-v[j].second, j});
        }
        lst = j;
      }
      res += cur_c;
      res2 += min(cur_c, -Set.begin()->first);
      Set.erase(Set.begin());
    }
    cout << res << ' ' << res2 << endl;
  }
  return 0;
}

```