#include <bits/stdc++.h>
const int maxn = 2 * (1e5 + 2);
using namespace std;
int q, a, b, n;
vector<int> types;
vector<int> good;
vector<int> v;
vector<int> rozmiary[maxn + 10];
vector<int> ilosci;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  types.resize(maxn + 10);
  good.resize(maxn + 10);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 0; i < n + 10; i++) {
      rozmiary[i].resize(0);
      good[i] = 0;
      types[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> a >> b;
      if (!b) good[a]++;
      types[a]++;
    }
    v.resize(0);
    ilosci.resize(0);
    for (int i = 1; i <= n; i++) {
      if (types[i]) {
        v.push_back(types[i]);
        rozmiary[types[i]].push_back(types[i] - good[i]);
      }
    }
    sort(v.begin(), v.end());
    int last = maxn;
    int ans1 = 0;
    int ans2 = 0;
    for (int i = v.size() - 1; i >= 0; i--) {
      if (v[i] >= last) {
        last--;
        ans1 += last;
        ilosci.push_back(last);
      } else {
        ans1 += v[i];
        last = v[i];
        ilosci.push_back(v[i]);
      }
      if (last == 1) break;
    }
    priority_queue<int> kolejka;
    for (int i = 0; i < ilosci.size(); i++) {
      for (int j = 0; j < rozmiary[ilosci[i]].size(); j++)
        kolejka.push(rozmiary[ilosci[i]][j]);
      ans2 += min(kolejka.top(), ilosci[i]);
      kolejka.pop();
    }
    cout << ans1 << " " << ans2 << "\n";
  }
}

### Prompt

Please provide a Cpp coded solution 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>
const int maxn = 2 * (1e5 + 2);
using namespace std;
int q, a, b, n;
vector<int> types;
vector<int> good;
vector<int> v;
vector<int> rozmiary[maxn + 10];
vector<int> ilosci;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  types.resize(maxn + 10);
  good.resize(maxn + 10);
  cin >> q;
  while (q--) {
    cin >> n;
    for (int i = 0; i < n + 10; i++) {
      rozmiary[i].resize(0);
      good[i] = 0;
      types[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> a >> b;
      if (!b) good[a]++;
      types[a]++;
    }
    v.resize(0);
    ilosci.resize(0);
    for (int i = 1; i <= n; i++) {
      if (types[i]) {
        v.push_back(types[i]);
        rozmiary[types[i]].push_back(types[i] - good[i]);
      }
    }
    sort(v.begin(), v.end());
    int last = maxn;
    int ans1 = 0;
    int ans2 = 0;
    for (int i = v.size() - 1; i >= 0; i--) {
      if (v[i] >= last) {
        last--;
        ans1 += last;
        ilosci.push_back(last);
      } else {
        ans1 += v[i];
        last = v[i];
        ilosci.push_back(v[i]);
      }
      if (last == 1) break;
    }
    priority_queue<int> kolejka;
    for (int i = 0; i < ilosci.size(); i++) {
      for (int j = 0; j < rozmiary[ilosci[i]].size(); j++)
        kolejka.push(rozmiary[ilosci[i]][j]);
      ans2 += min(kolejka.top(), ilosci[i]);
      kolejka.pop();
    }
    cout << ans1 << " " << ans2 << "\n";
  }
}

```