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;
pair<int, int> v[200005];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x, type;
      cin >> x >> type;
      v[x].first++;
      v[x].second += type;
    }
    sort(v + 1, v + n + 1);
    priority_queue<int> q;
    int crt = n;
    int l = n;
    int ans1 = 0, ans2 = 0;
    while (l > 0 && crt > 0) {
      while (v[l].first >= crt) {
        q.push(v[l].second);
        l--;
      }
      if (!q.empty()) {
        ans1 += crt;
        ans2 += min(crt, q.top());
        q.pop();
      }
      crt--;
    }
    for (int i = 1; i <= n; i++) v[i].first = v[i].second = 0;
    cout << ans1 << " " << ans2 << '\n';
  }
}