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;
const int MAX_N = 200002;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (0) freopen("Ginput.txt", "r", stdin);
  int q;
  cin >> q;
  for (int test_ = 0; test_ < (q); test_++) {
    int n;
    cin >> n;
    map<int, pair<int, int> > mp;
    for (int i = 0; i < (n); i++) {
      int a;
      bool give;
      cin >> a >> give;
      auto pr = mp[a];
      pr.first++;
      if (give) pr.second++;
      mp[a] = pr;
    }
    vector<pair<int, int> > cnts;
    cnts.reserve(mp.size());
    for (auto &pair : mp) {
      cnts.push_back(pair.second);
    }
    sort(cnts.begin(), cnts.end(),
         [](const pair<int, int> &a, const pair<int, int> &b) -> bool {
           if (a.first != b.first) return b.first < a.first;
           return b.second < a.second;
         });
    for (int i = 0; i < (cnts.size()); i++) {
      if (0)
        cerr << "cnts[i].first"
             << ": " << cnts[i].first << endl;
      ;
      if (0)
        cerr << "cnts[i].second"
             << ": " << cnts[i].second << endl;
      ;
    }
    if (0)
      cerr << "\"===\""
           << ": "
           << "===" << endl;
    ;
    long long sum = 0;
    long long sumf = 0;
    int cap = cnts[0].first;
    int i = 0;
    while (cap > 0 && i < cnts.size()) {
      cap = min(cap, cnts[i].first);
      int j = i;
      int max_se_j = i;
      int max_se = cnts[i].second;
      while (j + 1 < cnts.size() && cnts[j + 1].first >= cap) {
        j++;
        if (cnts[j].second > max_se) {
          max_se = cnts[j].second;
          max_se_j = j;
        }
      }
      swap(cnts[i], cnts[max_se_j]);
      if (0)
        cerr << "cnts[i].first"
             << ": " << cnts[i].first << endl;
      ;
      if (0)
        cerr << "cnts[i].second"
             << ": " << cnts[i].second << endl;
      ;
      sum += cap;
      sumf += min(cap, cnts[i].second);
      cap--;
      i++;
    }
    cout << sum << " " << sumf << endl;
  }
}