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;
void solve() {
  map<int, pair<int, int> > mapa;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a, f;
    cin >> a >> f;
    if (not mapa.count(a)) mapa[a] = pair<int, int>(0, 0);
    mapa[a].first++;
    mapa[a].second += f;
  }
  vector<pair<int, int> > v;
  for (map<int, pair<int, int> >::iterator it = mapa.begin(); it != mapa.end();
       it++)
    v.push_back(it->second);
  sort(v.begin(), v.end());
  int s = 0;
  int s1 = 0;
  priority_queue<int> q;
  int cur = 1000000000;
  for (int i = int(v.size()) - 1; i >= 0 and cur > 0;) {
    if (not(v[i].first < cur and int(q.size()))) {
      cur = v[i].first;
      for (; i >= 0 and v[i].first == cur; i--) q.push(v[i].second);
    }
    s += cur;
    s1 += min(cur, q.top());
    q.pop();
    cur--;
  }
  while (cur > 0 and int(q.size())) {
    s += cur;
    s1 += min(cur, q.top());
    q.pop();
    cur--;
  }
  cout << s << " " << s1 << endl;
}
int cases;
int main() {
  ios::sync_with_stdio(false);
  cin >> cases;
  for (int cas = 0; cas < cases; cas++) solve();
}