#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    int n, x, f, last = 1, ans = 0, profit = 0;
    vector<pair<long long, long long>> a;
    a.push_back({0, 0});
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> x >> f, a.push_back({x, f});
    sort(a.begin(), a.end());
    for (int i = 2; i <= n; i++) a[i].second += a[i - 1].second;
    priority_queue<pair<int, int>> q;
    for (int i = 2; i <= n; i++) {
      if (a[i].first != a[i - 1].first) {
        q.push({i - last, a[i - 1].second - a[last - 1].second});
        last = i;
      }
    }
    q.push({n - last + 1, a[n].second - a[last - 1].second});
    int upper = INT_MAX;
    ans += q.top().first;
    upper = q.top().first;
    profit += min(q.top().second, upper);
    q.pop();
    while (!q.empty() && upper > 1) {
      while (q.top().first > upper - 1) {
        int ff = q.top().first, ss = q.top().second;
        q.pop();
        q.push({ff - 1, min(ff - 1, ss)});
      }
      ans += min(q.top().first, upper - 1);
      upper = min(q.top().first, upper - 1);
      profit += min(q.top().second, upper);
      q.pop();
    }
    cout << ans << " " << profit << "\n";
  }
  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;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    int n, x, f, last = 1, ans = 0, profit = 0;
    vector<pair<long long, long long>> a;
    a.push_back({0, 0});
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> x >> f, a.push_back({x, f});
    sort(a.begin(), a.end());
    for (int i = 2; i <= n; i++) a[i].second += a[i - 1].second;
    priority_queue<pair<int, int>> q;
    for (int i = 2; i <= n; i++) {
      if (a[i].first != a[i - 1].first) {
        q.push({i - last, a[i - 1].second - a[last - 1].second});
        last = i;
      }
    }
    q.push({n - last + 1, a[n].second - a[last - 1].second});
    int upper = INT_MAX;
    ans += q.top().first;
    upper = q.top().first;
    profit += min(q.top().second, upper);
    q.pop();
    while (!q.empty() && upper > 1) {
      while (q.top().first > upper - 1) {
        int ff = q.top().first, ss = q.top().second;
        q.pop();
        q.push({ff - 1, min(ff - 1, ss)});
      }
      ans += min(q.top().first, upper - 1);
      upper = min(q.top().first, upper - 1);
      profit += min(q.top().second, upper);
      q.pop();
    }
    cout << ans << " " << profit << "\n";
  }
  return 0;
}

```