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() {
  long long int n, i, x, d, temp, sum = 0;
  cin >> n;
  multiset<long long int> am;
  set<long long int> ch;
  for (i = 0; i < n; i++) {
    cin >> x;
    am.insert(x);
    ch.insert(x);
  }
  x = ch.size();
  multiset<long long int>::iterator itr;
  itr = am.begin();
  long long int a[x];
  for (i = 0; i < x; i++) {
    d = am.count(*itr);
    a[i] = d;
    am.erase(*itr);
    itr = am.begin();
  }
  sort(a, a + x, greater<long long int>());
  temp = a[0];
  sum += temp;
  for (i = 1; i < x; i++) {
    if (a[i] == temp) {
      sum += (a[i] - 1);
      temp = a[i] - 1;
    } else if (a[i] > temp) {
      sum += (temp - 1);
      temp = (temp - 1);
    } else {
      sum += a[i];
      temp = a[i];
    }
    if (temp == 1 || temp <= 0) break;
  }
  cout << sum << endl;
}
int main() {
  long long int q;
  cin >> q;
  while (q--) solve();
  return 0;
}