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;
long long n, a[202020], cnt[202020], mx, f[202020];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  cin >> q;
  while (q--) {
    cin >> n;
    set<long long> s;
    multiset<pair<long long, long long>> t;
    long long ans = 0, fans = 0;
    mx = 0;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> a[i] >> x;
      if (x == 0) {
        f[a[i]]++;
      }
      s.insert(a[i]);
      cnt[a[i]]++;
      mx = max(mx, cnt[a[i]]);
    }
    for (auto u : s) {
      t.insert({cnt[u], f[u]});
    }
    for (long long i = mx; i >= 1; i--) {
      pair<long long, long long> p = {0, 0};
      long long cur = 1e18;
      for (auto u : t) {
        if ((u).first >= i) {
          long long dif = max(1LL * 0, i - ((u.first) - (u.second)));
          if (dif <= cur) {
            p = u;
            cur = dif;
          }
        }
      }
      if (p.first != 0) {
        ans += i;
        t.erase(t.find(p));
        fans += cur;
      }
    }
    cout << ans << " " << ans - fans << endl;
    for (auto u : s) {
      cnt[u] = 0;
      f[u] = 0;
    }
    t.clear();
    s.clear();
  }
  return 0;
}