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 sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
int main() {
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    long long a[n];
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mp[a[i]]++;
    }
    vector<pair<int, int>> v;
    for (auto it : mp) {
      v.push_back(make_pair(it.first, it.second));
    }
    sort(v.begin(), v.end(), sortbysec);
    long long res = 0;
    res += v[0].second;
    long long temp = v[0].second;
    for (int i = 1; i < v.size(); i++) {
      temp--;
      if (temp == 0) {
        break;
      }
      long long d = (v[i].second - temp);
      if (d >= 0) {
        res += temp;
      } else {
        res += v[i].second;
        temp = v[i].second;
      }
    }
    cout << res << endl;
  }
  return 0;
}