This problem is a version of problem D from the same contest with some additional constraints and tasks.

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.

You really like some of the candies and don't want to include them into the gift, but you want to eat them yourself instead. For each candy, a number f_i is given, which is equal to 0 if you really want to keep i-th candy for yourself, or 1 if you don't mind including it into your gift. It is possible that two candies of the same type have different values of f_i.

You want your gift to be as large as possible, but you don't want to include too many of the candies you want to eat into the gift. So, you want to calculate the maximum possible number of candies that can be included into a gift, and among all ways to choose maximum number of candies, you want to maximize the number of candies having f_i = 1 in your gift.

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.

The first line of each query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of candies.

Then n lines follow, each containing two integers a_i and f_i (1 ≤ a_i ≤ n, 0 ≤ f_i ≤ 1), where a_i is the type of the i-th candy, and f_i denotes whether you want to keep the i-th candy for yourself (0 if you want to keep it, 1 if you don't mind giving it away).

It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.

Output

For each query print two integers:

  * the maximum number of candies in a gift you can compose, according to the constraints in the statement; 
  * the maximum number of candies having f_i = 1 in a gift you can compose that contains the maximum possible number of candies. 

Example

Input


3
8
1 0
4 1
2 0
4 1
5 1
6 1
3 0
2 0
4
1 1
1 1
2 1
2 1
9
2 0
2 0
4 1
4 1
4 1
7 0
7 1
7 0
7 1


Output


3 3
3 3
9 5

Note

In the first query, you can include two candies of type 4 and one candy of type 5. All of them have f_i = 1 and you don't mind giving them away as part of the gift.
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  for (; q > 0; q--) {
    int n;
    cin >> n;
    vector<pair<int, int>> a(n);
    for (int i = 0; i < n; i++) {
      int ai, fi;
      cin >> ai >> fi;
      ai--;
      a[ai].first += 1;
      if (fi == 1) a[ai].second += 1;
    }
    sort(a.begin(), a.end(), greater<pair<int, int>>());
    vector<int> sol;
    for (int i = 0, mm = INF; i < n; i++) {
      if (min(mm, a[i].first) > 0)
        sol.push_back(min(mm, a[i].first));
      else
        break;
      mm = max(0, sol[i] - 1);
    }
    int ans = 0;
    for (int i = 0; i < (int)sol.size(); i++) {
      ans += sol[i];
    }
    cout << ans << " ";
    multiset<int, greater<int>> st;
    vector<int> nsol;
    for (int i = 0, j = 0; i < (int)sol.size(); i++) {
      for (; j < n && a[j].first >= sol[i]; j++) {
        st.insert(a[j].second);
      }
      auto it = st.begin();
      nsol.push_back(min(*(it), sol[i]));
      st.erase(it);
    }
    ans = 0;
    for (int i = 0; i < (int)nsol.size(); i++) {
      ans += nsol[i];
    }
    cout << ans << endl;
  }
  return 0;
}
