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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int test_cases;
  cin >> test_cases;
  while (test_cases--) {
    long long int n;
    cin >> n;
    vector<long long int> v(n);
    for (long long int i = 0; i < n; i++) cin >> v[i];
    map<long long int, long long int> m;
    for (long long int i = 0; i <= n - 1; i++) {
      m[v[i]]++;
    }
    vector<pair<long long int, long long int> > vvv;
    for (auto i : m) {
      vvv.push_back({i.second, i.first});
    }
    sort(vvv.begin(), vvv.end());
    long long int index = vvv.size() - 1;
    pair<long long int, long long int> p = vvv[index];
    long long int cur = p.first;
    long long int ans = 0;
    while (cur > 0) {
      ans += cur;
      index--;
      if (index >= 0) {
        pair<long long int, long long int> q = vvv[index];
        cur = min(cur - 1, q.first);
      } else
        break;
    }
    std::cout << ans << std::endl;
  }
  return 0;
}