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 int m = 1000000007;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first)
    return true;
  else if (a.first == b.first && a.first + a.second > b.first + b.second)
    return true;
  return false;
}
int main() {
  int t;
  cin >> t;
  for (int q = 0; q < t; q++) {
    int n;
    cin >> n;
    map<int, pair<int, int>> mp;
    for (int i = 0; i < n; i++) {
      int elem, f;
      cin >> elem >> f;
      if (f == 1)
        mp[elem].first++;
      else
        mp[elem].second++;
    }
    int len = mp.size();
    pair<int, int> arr[len];
    int ind = 0;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
      arr[ind] = itr->second;
      ind++;
    }
    sort(arr, arr + len, cmp);
    long long int ans = 0;
    unordered_map<int, int> freqs;
    int tot = 0;
    for (int i = 0; i < len; i++) {
      int freq = arr[i].first + arr[i].second;
      if (freqs.find(freq) == freqs.end()) {
        ans += freq;
        freqs[freq]++;
        tot += arr[i].first;
      } else {
        int zero = arr[i].second;
        while (freqs.find(freq) != freqs.end() && zero > 0) {
          freq--;
          zero--;
        }
        int one = arr[i].first;
        while (freqs.find(freq) != freqs.end() && one > 0) {
          freq--;
          one--;
        }
        tot += one;
        ans += freq;
        freqs[freq]++;
      }
    }
    cout << ans << " " << tot << endl;
  }
}