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 n;
int main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<int> num(n + 1), num_good(n + 1);
    for (int i = (1); i <= (n); ++i) {
      int aa, xx;
      cin >> aa >> xx;
      num[aa]++;
      if (xx) num_good[aa]++;
    }
    vector<vector<int> > type(n + 1);
    for (int i = (1); i <= (n); ++i) {
      type[num[i]].push_back(num_good[i]);
    }
    int ans1 = 0;
    int ans2 = 0;
    priority_queue<int> q;
    for (int i = (n); i >= (1); --i) {
      for (auto x : type[i]) q.push(x);
      if (!q.empty()) {
        int max_val = q.top();
        q.pop();
        ans1 += i;
        ans2 += min(i, max_val);
      }
    }
    printf("%d %d\n", ans1, ans2);
  }
}