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;
using ll = long long;
using ld = long double;
struct Candy {
  int nr;
  int g;
};
bool cmp(Candy a, Candy b) { return a.nr > b.nr; }
bool operator<(Candy a, Candy b) { return a.g < b.g; }
const int MAX_N = 2e5;
Candy f[1 + MAX_N];
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    f[x].nr++;
    if (y) f[x].g++;
  }
  int ans = 0, ansf = 0;
  sort(f + 1, f + n + 1, cmp);
  priority_queue<Candy> pq;
  pq.push(f[1]);
  int mn = f[1].nr;
  int i = 2;
  while (mn) {
    while (i <= n && f[i].nr == mn) {
      pq.push(f[i]);
      i++;
    }
    if (!pq.empty()) {
      Candy kek = pq.top();
      ans += mn;
      ansf += min(mn, kek.g);
      pq.pop();
    }
    mn--;
  }
  cout << ans << " " << ansf << "\n";
  for (int i = 1; i <= n; i++) f[i] = {0, 0};
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) solve();
}