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;
const int MAX = 2e5 + 3, K = 20;
const long long INF = 1e18;
void solve() {
  int n;
  cin >> n;
  vector<int> cnt(n + 1, 0), cnt2(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    int a, f;
    cin >> a >> f;
    cnt[a]++;
    cnt2[a] += f;
  }
  vector<pair<int, int> > v;
  for (int i = 1; i <= n; i++) {
    v.push_back({cnt[i], cnt2[i]});
  }
  sort(v.rbegin(), v.rend());
  int j = n;
  int mx1 = 0, mx2 = 0;
  set<int> take;
  for (int i = 0; i < n and j > 0; i++) {
    while (j > v[i].first) j--;
    mx1 += j;
    take.insert(j);
    j--;
  }
  for (int i = 0; i < n; i++) swap(v[i].first, v[i].second);
  sort(v.rbegin(), v.rend());
  for (int i = 0; i < n and !take.empty(); i++) {
    auto it = take.upper_bound(v[i].second);
    if (it != take.begin()) {
      it--;
      mx2 += min(v[i].first, *it);
      take.erase(it);
    }
  }
  cout << mx1 << " " << mx2 << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}