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 main() {
  long long int n, i, j, k, x, y, t, m, cnt;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<pair<long long int, long long int> > V;
    vector<pair<long long int, long long int> > tmp;
    priority_queue<pair<long long int, long long int> > Q;
    for (i = 1; i <= n; i++) {
      cin >> x >> y;
      V.push_back(make_pair(x, y));
    }
    sort(V.begin(), V.end());
    x = V.size();
    i = 0, j = 0;
    while (i < x) {
      j = i;
      long long int f = 0;
      while (j < x) {
        if (V[j].first != V[i].first) break;
        if (V[j].second == 1) f++;
        j++;
      }
      Q.push(make_pair(j - i, f));
      i = j;
    }
    long long int ans = 0, an = 0;
    while (!Q.empty()) {
      pair<long long int, long long int> P = Q.top();
      Q.pop();
      ans += P.first, an += min(P.second, P.first);
      while (!Q.empty() && Q.top().first == P.first) {
        pair<long long int, long long int> T = Q.top();
        Q.pop();
        T.first--;
        if (T.first) Q.push(T);
      }
    }
    cout << ans << " " << an << endl;
  }
}