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 A[200005], F[200005];
vector<int> S[200005];
void pro() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) A[i] = F[i] = 0, S[i].clear();
  for (int i = 0; i < n; i++) {
    int a, f;
    cin >> a >> f;
    A[a]++;
    F[a] += f;
  }
  for (int i = 1; i <= n; i++) {
    S[A[i]].push_back(F[i]);
  }
  priority_queue<int> pq;
  int acnt = 0, fcnt = 0;
  for (int i = n; i > 0; i--) {
    for (int j : S[i]) pq.push(j);
    if (pq.empty()) continue;
    acnt += i;
    fcnt += min(i, pq.top());
    pq.pop();
  }
  cout << acnt << ' ' << fcnt << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) pro();
  return 0;
}