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 power(long long x, long long y);
long long gcd(long long a, long long b);
int sum(int x) {
  int s = 0;
  while (x > 0) {
    s += (x % 10);
    x /= 10;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    map<int, pair<int, int>> mp;
    for (int i = 0; i < n; i++) {
      int val, type;
      cin >> val >> type;
      if (type)
        mp[val].second++;
      else
        mp[val].first++;
    }
    auto it = mp.begin();
    priority_queue<pair<int, int>> pq;
    priority_queue<pair<int, int>> pq1;
    while (it != mp.end()) {
      pq.push({it->second.first + it->second.second, it->first});
      it++;
    }
    int curr = pq.top().first;
    int ans = 0;
    int ones = 0;
    pq1.push({mp[pq.top().second].second, mp[pq.top().second].first});
    pq.pop();
    while (curr > 0) {
      if (pq.empty() && pq1.empty()) break;
      while (!pq.empty() && pq.top().first >= curr) {
        pq1.push({mp[pq.top().second].second, mp[pq.top().second].first});
        pq.pop();
      }
      if (pq1.empty()) {
        curr--;
        continue;
      }
      if (curr > (pq1.top().first + pq1.top().second)) {
        curr = pq1.top().first + pq1.top().second - 1;
        ans += pq1.top().first + pq1.top().second;
        ones += pq1.top().first;
      } else {
        if (curr <= pq1.top().first) {
          ans += curr;
          ones += curr;
        } else {
          ones += pq1.top().first;
          ans += curr;
        }
        curr--;
      }
      pq1.pop();
    }
    cout << ans << " " << ones << "\n";
  }
  return 0;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}