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;
const long long MOD = 1e9 + 7;
const long long N = 2e5 + 5;
const long double pi = 3.14159265359;
multiset<long long, greater<long long> > s;
long long T, n, x, f, ansa, ansb, cnt[2][N];
vector<long long> v[N];
void reset(long long k) {
  for (long long i = 0; i < k + 1; i++) {
    cnt[0][i] = cnt[1][i] = 0;
    v[i].clear();
  }
  s.clear();
  ansa = 0;
  ansb = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> x >> f;
      cnt[f][x]++;
    }
    for (long long i = 1; i <= n; i++) {
      v[cnt[0][i] + cnt[1][i]].push_back(cnt[1][i]);
    }
    for (long long i = n; i >= 1; i--) {
      for (auto ii : v[i]) s.insert(ii);
      if (s.size()) {
        ansa += i;
        ansb += min(*s.begin(), i);
        s.erase(s.begin());
      }
    }
    cout << ansa << " " << ansb << "\n";
    reset(n);
  }
  return 0;
}