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;
signed main() {
  long long t;
  cin >> t;
  for (long long q = 0; q < t; q++) {
    long long n;
    cin >> n;
    long long mass[n];
    for (long long i = 0; i < n; i++) cin >> mass[i];
    sort(mass, mass + n);
    vector<long long> ans;
    long long c = 1;
    for (long long i = 1; i < n; i++) {
      if (mass[i] == mass[i - 1]) {
        c++;
      } else {
        ans.push_back(c);
        c = 1;
      }
    }
    ans.push_back(c);
    sort(ans.begin(), ans.end());
    reverse(ans.begin(), ans.end());
    set<long long> perm;
    set<long long> res;
    for (long long i = 0; i < ans[0]; i++) {
      perm.insert(i + 1);
    }
    for (long long z = 0; z < ans.size(); z++) {
      long long x = ans[z];
      set<long long>::iterator it = perm.lower_bound(x);
      if (*it == x) {
        res.insert(x);
        perm.erase(x);
        continue;
      }
      if (it == perm.begin()) {
        continue;
      } else {
        --it;
        x = *it;
        perm.erase(x);
        res.insert(x);
      }
    }
    long long flex = 0;
    for (set<long long>::iterator it = res.begin(); it != res.end(); it++) {
      flex += *it;
    }
    cout << flex << endl;
  }
}