#include <bits/stdc++.h>
using namespace std;
struct te {
  long a, b, c;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long q;
  cin >> q;
  while (q--) {
    long n, i = 0, prev = 0, ans = 0, p = 0, f, a, b, c, val = 0, maxm = 0;
    cin >> n;
    map<long, long[2]> mp;
    map<long, long> mp1;
    map<long, long> v[n + 1];
    for (i = 0; i < n; i++) {
      cin >> a >> f;
      if (mp.find(a) != mp.end()) {
        if (f == 0)
          mp[a][0]++;
        else
          mp[a][1]++;
      } else {
        if (f == 0) {
          mp[a][0] = 1;
          mp[a][1] = 0;
        } else {
          mp[a][0] = 0;
          mp[a][1] = 1;
        }
      }
    }
    for (auto it : mp) {
      b = mp[it.first][0];
      c = mp[it.first][1];
      if (v[b + c].find(c) != v[b + c].end())
        v[b + c][c]++;
      else
        v[b + c][c] = 1;
      maxm = max(maxm, b + c);
    }
    for (i = maxm; i >= 1; i--) {
      if (v[i].size() > 0) {
        for (auto it : v[i]) {
          if (mp1.find(it.first) != mp1.end())
            mp1[it.first] += it.second;
          else
            mp1[it.first] = it.second;
        }
      }
      if (mp1.size() > 0) {
        ans += i;
        auto it = --mp1.end();
        val += min(it->first, i);
        it->second = it->second - 1;
        if (it->second == 0) mp1.erase(it);
      }
    }
    cout << ans << " " << val << endl;
  }
}

### Prompt

Please formulate 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;
struct te {
  long a, b, c;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long q;
  cin >> q;
  while (q--) {
    long n, i = 0, prev = 0, ans = 0, p = 0, f, a, b, c, val = 0, maxm = 0;
    cin >> n;
    map<long, long[2]> mp;
    map<long, long> mp1;
    map<long, long> v[n + 1];
    for (i = 0; i < n; i++) {
      cin >> a >> f;
      if (mp.find(a) != mp.end()) {
        if (f == 0)
          mp[a][0]++;
        else
          mp[a][1]++;
      } else {
        if (f == 0) {
          mp[a][0] = 1;
          mp[a][1] = 0;
        } else {
          mp[a][0] = 0;
          mp[a][1] = 1;
        }
      }
    }
    for (auto it : mp) {
      b = mp[it.first][0];
      c = mp[it.first][1];
      if (v[b + c].find(c) != v[b + c].end())
        v[b + c][c]++;
      else
        v[b + c][c] = 1;
      maxm = max(maxm, b + c);
    }
    for (i = maxm; i >= 1; i--) {
      if (v[i].size() > 0) {
        for (auto it : v[i]) {
          if (mp1.find(it.first) != mp1.end())
            mp1[it.first] += it.second;
          else
            mp1[it.first] = it.second;
        }
      }
      if (mp1.size() > 0) {
        ans += i;
        auto it = --mp1.end();
        val += min(it->first, i);
        it->second = it->second - 1;
        if (it->second == 0) mp1.erase(it);
      }
    }
    cout << ans << " " << val << endl;
  }
}

```