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;
typedef vector<vector<int> > VVI;
typedef priority_queue<int, vector<int>, greater<int> > PQ;
bool ok(vector<int> V, int k, int x) {
  for (int y : V) {
    int df = abs(x - y);
    if (df > k) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    vector<int> V, V2;
    vector<pair<int, int> > Ar;
    unordered_map<int, int> Map, Map2, Map3;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      V.push_back(x);
      Map[x]++;
    }
    for (int i = 0; i < n; i++) {
      if (Map3[V[i]] == 0) {
        Map3[V[i]] = 1;
        Ar.push_back({Map[V[i]], V[i]});
      }
    }
    sort(Ar.rbegin(), Ar.rend());
    int last = 1e9;
    for (int i = 0; i < Ar.size(); i++) {
      int x = Ar[i].second;
      int f = Map[Ar[i].second];
      if (!Map2[f]) {
        V2.push_back(f);
        Map2[f] = 1;
        last = f;
      } else {
        if (last > 1) {
          last--;
          V2.push_back(last);
          Map2[last] = 1;
        }
      }
    }
    int ans = accumulate(V2.begin(), V2.end(), 0);
    cout << ans << "\n";
  }
  return 0;
}