#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int i;
    vector<pair<long long int, long long int> > v;
    set<long long int> broj;
    priority_queue<long long int> q;
    map<long long int, long long int> m, m1;
    for (i = 0; i < n; i++) {
      long long int a, b;
      cin >> a >> b;
      m[a]++;
      m1[a] += b;
    }
    for (auto it : m) {
      broj.insert(it.second);
      v.push_back({it.second, m1[it.first]});
    }
    sort(v.begin(), v.end());
    long long int ans = 0, ans1 = 0;
    while (broj.size()) {
      long long int x = *(--broj.end());
      broj.erase(--broj.end());
      while (v.size() && v.back().first >= x) {
        q.push(v.back().second);
        v.pop_back();
      }
      long long int pos = 0;
      if (broj.size() >= 1) pos = *(--broj.end());
      while (x > pos && q.size() > 0) {
        long long int y = q.top();
        ans += min(x, y);
        ans1 += x;
        q.pop();
        x--;
      }
    }
    cout << ans1 << " " << ans << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int i;
    vector<pair<long long int, long long int> > v;
    set<long long int> broj;
    priority_queue<long long int> q;
    map<long long int, long long int> m, m1;
    for (i = 0; i < n; i++) {
      long long int a, b;
      cin >> a >> b;
      m[a]++;
      m1[a] += b;
    }
    for (auto it : m) {
      broj.insert(it.second);
      v.push_back({it.second, m1[it.first]});
    }
    sort(v.begin(), v.end());
    long long int ans = 0, ans1 = 0;
    while (broj.size()) {
      long long int x = *(--broj.end());
      broj.erase(--broj.end());
      while (v.size() && v.back().first >= x) {
        q.push(v.back().second);
        v.pop_back();
      }
      long long int pos = 0;
      if (broj.size() >= 1) pos = *(--broj.end());
      while (x > pos && q.size() > 0) {
        long long int y = q.top();
        ans += min(x, y);
        ans1 += x;
        q.pop();
        x--;
      }
    }
    cout << ans1 << " " << ans << endl;
  }
  return 0;
}

```