#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long bin(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  if (b & 1) return (a * bin((a * a) % mod, b / 2, mod)) % mod;
  return (1 * bin((a * a) % mod, b / 2, mod)) % mod;
}
void I_m_Beast() {
  int n, a, b;
  cin >> n;
  int A[n + 1][2];
  memset(A, 0, sizeof(A));
  vector<int> v[n + 1];
  for (int i = 1; i < n + 1; i++) {
    cin >> a >> b;
    A[a][0]++;
    A[a][1] += b;
  }
  for (int i = 1; i < n + 1; i++) {
    if (A[i][0]) v[A[i][0]].push_back(A[i][1]);
  }
  for (int i = 1; i < n + 1; i++) sort(v[i].begin(), v[i].end());
  int ans = 0, left = 0, ansf = 0;
  for (int i = n; i >= 1; i--) {
    if (v[i].size()) {
      ans += i;
      int x = (int)v[i].size();
      ansf += v[i][x - 1];
      v[i].pop_back();
      for (int j = 0; j < (int)v[i].size(); j++) {
        v[i - 1].push_back(min(i - 1, v[i][j]));
      }
      sort(v[i - 1].begin(), v[i - 1].end());
    }
  }
  cout << ans << " " << ansf << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    I_m_Beast();
  }
  return 0;
}

### 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;
const long long mod = 1e9 + 7;
long long bin(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  if (b & 1) return (a * bin((a * a) % mod, b / 2, mod)) % mod;
  return (1 * bin((a * a) % mod, b / 2, mod)) % mod;
}
void I_m_Beast() {
  int n, a, b;
  cin >> n;
  int A[n + 1][2];
  memset(A, 0, sizeof(A));
  vector<int> v[n + 1];
  for (int i = 1; i < n + 1; i++) {
    cin >> a >> b;
    A[a][0]++;
    A[a][1] += b;
  }
  for (int i = 1; i < n + 1; i++) {
    if (A[i][0]) v[A[i][0]].push_back(A[i][1]);
  }
  for (int i = 1; i < n + 1; i++) sort(v[i].begin(), v[i].end());
  int ans = 0, left = 0, ansf = 0;
  for (int i = n; i >= 1; i--) {
    if (v[i].size()) {
      ans += i;
      int x = (int)v[i].size();
      ansf += v[i][x - 1];
      v[i].pop_back();
      for (int j = 0; j < (int)v[i].size(); j++) {
        v[i - 1].push_back(min(i - 1, v[i][j]));
      }
      sort(v[i - 1].begin(), v[i - 1].end());
    }
  }
  cout << ans << " " << ansf << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    I_m_Beast();
  }
  return 0;
}

```