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 long long Mod = 1e9 + 7;
const double pi = 2 * acos(0.0);
const int N = 2e5 + 5;
int main() {
  int q;
  scanf("%d", &(q));
  while (q--) {
    int n;
    scanf("%d", &(n));
    map<int, int> freq, counts;
    set<int> free;
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &(x));
      freq[x]++;
    }
    for (auto& it : freq) {
      counts[it.second]++;
    }
    for (int i = 1; i <= n; i++) {
      if (!counts.count(i)) free.insert(i);
    }
    for (auto it = freq.begin(); it != freq.end();) {
      bool inc = true;
      if (counts.count(it->second) && counts[it->second] > 1) {
        if (!free.empty()) {
          auto num = free.lower_bound(it->second);
          if (*num > it->second && num != free.begin()) num--;
          if (*num < it->second) {
            counts[it->second]--;
            freq[it->first] = *num;
            counts[it->second]++;
            free.erase(num);
          } else if (*num > it->second && num == free.begin()) {
            counts[it->second]--;
            it = freq.erase(it);
            inc = false;
          }
        } else {
          counts[it->second]--;
          it = freq.erase(it);
          inc = false;
        }
      }
      if (inc) it++;
    }
    int ans = 0;
    for (auto& it : freq) {
      ans += it.second;
    }
    printf("%d\n", ans);
  }
  return 0;
}