#include <bits/stdc++.h>
using namespace std;
void solve();
void YN(bool b) {
  if (b) {
    cout << "YES"
         << "\n";
    ;
  } else {
    cout << "NO"
         << "\n";
    ;
  }
};
void yn(bool b) {
  if (b) {
    cout << "Yes"
         << "\n";
    ;
  } else {
    cout << "No"
         << "\n";
    ;
  }
};
void queries() {
  int n;
  cin >> n;
  while (n--) solve();
}
const int maxn = 200200;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  for (auto i : m) {
    b.push_back(i.second);
  }
  map<int, bool> taken;
  long long sum = 0;
  sort(b.begin(), b.end());
  for (int i = b.size() - 1; i >= 0; i--) {
    if (!taken[b[i]]) {
      sum += b[i];
      taken[b[i]] = 1;
    } else {
      for (int j = b[i]; j > 0; j--) {
        if (!taken[j]) {
          sum += j;
          taken[j] = 1;
          break;
        }
      }
    }
  }
  cout << sum << "\n";
  ;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  queries();
  ;
  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;
void solve();
void YN(bool b) {
  if (b) {
    cout << "YES"
         << "\n";
    ;
  } else {
    cout << "NO"
         << "\n";
    ;
  }
};
void yn(bool b) {
  if (b) {
    cout << "Yes"
         << "\n";
    ;
  } else {
    cout << "No"
         << "\n";
    ;
  }
};
void queries() {
  int n;
  cin >> n;
  while (n--) solve();
}
const int maxn = 200200;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  for (auto i : m) {
    b.push_back(i.second);
  }
  map<int, bool> taken;
  long long sum = 0;
  sort(b.begin(), b.end());
  for (int i = b.size() - 1; i >= 0; i--) {
    if (!taken[b[i]]) {
      sum += b[i];
      taken[b[i]] = 1;
    } else {
      for (int j = b[i]; j > 0; j--) {
        if (!taken[j]) {
          sum += j;
          taken[j] = 1;
          break;
        }
      }
    }
  }
  cout << sum << "\n";
  ;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  queries();
  ;
  return 0;
}

```