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;
long long q, n, x, pri, pre, ans, ac;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  while (q--) {
    cin >> n;
    set<long long> s, p;
    vector<long long> cnt(n + 1), cntpri(n + 1), next(n + 1);
    vector<vector<long long>> e(n + 1);
    for (int i = 0; i < n; i++) {
      cin >> x >> pri;
      s.insert(x);
      cnt[x]++;
      cntpri[x] += pri;
    }
    for (int u : s) {
      e[cnt[u]].push_back(cntpri[u]);
      p.insert(cnt[u]);
    }
    vector<long long> v(p.begin(), p.end());
    for (int i = (int)v.size() - 1; i > 0; i--) next[v[i]] = v[i - 1];
    ans = 0, ac = 0;
    for (int i = (int)v.size() - 1; i >= 0; i--) {
      if (e[v[i]].size() > v[i] - next[v[i]]) {
        sort(e[v[i]].begin(), e[v[i]].end());
        reverse(e[v[i]].begin(), e[v[i]].end());
        while (e[v[i]].size() > v[i] - next[v[i]]) {
          e[next[v[i]]].push_back(e[v[i]].back());
          e[v[i]].pop_back();
        }
      }
      sort(e[v[i]].begin(), e[v[i]].end());
      reverse(e[v[i]].begin(), e[v[i]].end());
      pre = v[i];
      for (int j = 0; j < e[v[i]].size(); j++) {
        ans += pre;
        if (e[v[i]][j] > pre)
          ac += pre;
        else
          ac += e[v[i]][j];
        pre--;
      }
    }
    cout << ans << " " << ac << "\n";
  }
  return 0;
}