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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int q;
  cin >> q;
  while (q--) {
    long long int n, s = 0, x, ans = 0;
    cin >> n;
    map<long long int, long long int> mp, ct;
    map<long long int, pair<long long int, long long int> > mpmp;
    vector<long long int> v(n);
    pair<long long int, long long int> p;
    vector<pair<long long int, long long int> > vp;
    for (long long int i = 1; i <= n; i++) {
      mpmp[i] = {0, 0};
      mp[i] = 0;
      ct[i] = 0;
    }
    for (long long int i = 0; i < n; i++) {
      cin >> v[i] >> x;
      mp[v[i]]++;
      if (x) ct[v[i]]++;
    }
    map<long long int, long long int>::iterator it = mp.begin();
    for (; it != mp.end(); it++) {
      long long int nr = it->second;
      long long int no = it->first;
      long long int ty = ct[no];
      if (mpmp[nr].first == 0)
        mpmp[nr] = {1, ty};
      else {
        p = mpmp[nr];
        while (nr > 0) {
          if (p.first == 0) {
            mpmp[nr] = {1, ty};
            break;
          }
          if (ty > p.second) {
            swap(ty, p.second);
            mpmp[nr] = {1, p.second};
          }
          nr--;
          p = mpmp[nr];
        }
      }
    }
    for (long long int i = 1; i <= n; i++)
      if (mpmp[i].first) {
        s += i;
        ans += min(i, mpmp[i].second);
      }
    cout << s << " " << ans << "\n";
  }
}