#include <bits/stdc++.h>
using namespace std;
int main() {
  long long T;
  cin >> T;
  for (int i = 0; i < T; i++) {
    long long n;
    cin >> n;
    long long arr[n];
    for (int j = 0; j < n; j++) cin >> arr[j];
    map<int, int> mp;
    for (int j = 0; j < n; j++) {
      if (mp.find(arr[j]) != mp.end())
        mp[arr[j]]++;
      else {
        mp.insert(make_pair(arr[j], 1));
      }
    }
    priority_queue<int> q;
    for (auto j = mp.begin(); j != mp.end(); j++) {
      q.push(j->second);
    }
    long long sum = 0;
    long long prev = -1;
    queue<int> count;
    while (!q.empty()) {
      long long temp = q.top();
      q.pop();
      if (prev == -1)
        sum = sum + temp;
      else {
        if (temp == prev) {
          count.push(temp);
          continue;
        } else if (temp < prev - 1) {
          while (!count.empty() && temp != prev - 1) {
            sum = sum + prev - 1;
            count.pop();
            prev = prev - 1;
          }
        }
        sum = sum + temp;
      }
      prev = temp;
    }
    prev = prev - 1;
    while (!count.empty() && prev > 0) {
      sum = sum + prev;
      count.pop();
      prev = prev - 1;
    }
    cout << sum << endl;
  }
}

### Prompt

Your task is to create 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;
  for (int i = 0; i < T; i++) {
    long long n;
    cin >> n;
    long long arr[n];
    for (int j = 0; j < n; j++) cin >> arr[j];
    map<int, int> mp;
    for (int j = 0; j < n; j++) {
      if (mp.find(arr[j]) != mp.end())
        mp[arr[j]]++;
      else {
        mp.insert(make_pair(arr[j], 1));
      }
    }
    priority_queue<int> q;
    for (auto j = mp.begin(); j != mp.end(); j++) {
      q.push(j->second);
    }
    long long sum = 0;
    long long prev = -1;
    queue<int> count;
    while (!q.empty()) {
      long long temp = q.top();
      q.pop();
      if (prev == -1)
        sum = sum + temp;
      else {
        if (temp == prev) {
          count.push(temp);
          continue;
        } else if (temp < prev - 1) {
          while (!count.empty() && temp != prev - 1) {
            sum = sum + prev - 1;
            count.pop();
            prev = prev - 1;
          }
        }
        sum = sum + temp;
      }
      prev = temp;
    }
    prev = prev - 1;
    while (!count.empty() && prev > 0) {
      sum = sum + prev;
      count.pop();
      prev = prev - 1;
    }
    cout << sum << endl;
  }
}

```