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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool myfunc(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) {
    return a.second > b.second;
  } else
    return a.first > b.first;
}
void move_down(unordered_map<int, int>& dir, int f, int s) {
  if (f <= 0) {
    return;
  }
  while (((dir).find(f) != (dir).end()) && dir[f] >= s) {
    f--;
  }
  if (((dir).find(f) != (dir).end())) {
    move_down(dir, f, dir[f]);
  }
  s = min(f, s);
  if (f > 0) dir[f] = s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS;
  cin >> TESTS;
  while (TESTS--) {
    int n;
    cin >> n;
    pair<int, int> A[n];
    memset(A, 0, sizeof(A));
    for (int i = 0; i < n; i++) {
      int j, k;
      cin >> j >> k;
      A[j - 1].first++;
      if (k == 1) {
        A[j - 1].second++;
      }
    }
    sort(A, A + n, myfunc);
    unordered_map<int, int> dir;
    for (int i = 0; i < n; i++) {
      int f = A[i].first;
      int s = A[i].second;
      if (f) {
        if (((dir).find(f) != (dir).end())) {
          int s2 = dir[f];
          if (s2 >= s) {
            move_down(dir, f, s);
          } else {
            dir[f] = s;
            move_down(dir, f, s2);
          }
        } else {
          dir[f] = s;
        }
      }
    }
    int ans1 = 0;
    int ans2 = 0;
    for (auto ptr = dir.begin(); ptr != dir.end(); ptr++) {
      ans1 += ptr->first;
      ans2 += ptr->second;
    }
    cout << ans1 << " " << ans2 << '\n';
  }
  return 0;
}