#include <bits/stdc++.h>
using namespace std;
long long int INF = 1e9 + 5;
long long int mod = 998244353;
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> mp;
  vector<long long int> v;
  long long int i, in;
  for (i = 0; i < n; i++) {
    cin >> in;
    mp[in]++;
  }
  for (auto ele : mp) {
    v.push_back(ele.second);
  }
  sort(v.begin(), v.end());
  queue<long long int> q;
  long long int prev = INF, count = INF, ans = 0;
  for (i = v.size() - 1; i >= 0; i--) {
    while (count > v[i] && !q.empty()) {
      long long int ele = q.front();
      q.pop();
      ans += min(count, ele);
      count--;
    }
    if (v[i] > count)
      q.push(v[i]);
    else {
      ans += v[i];
      count = v[i] - 1;
    }
  }
  while (!q.empty() && count > 0) {
    ans += count;
    count--;
    q.pop();
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in CPP to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int INF = 1e9 + 5;
long long int mod = 998244353;
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> mp;
  vector<long long int> v;
  long long int i, in;
  for (i = 0; i < n; i++) {
    cin >> in;
    mp[in]++;
  }
  for (auto ele : mp) {
    v.push_back(ele.second);
  }
  sort(v.begin(), v.end());
  queue<long long int> q;
  long long int prev = INF, count = INF, ans = 0;
  for (i = v.size() - 1; i >= 0; i--) {
    while (count > v[i] && !q.empty()) {
      long long int ele = q.front();
      q.pop();
      ans += min(count, ele);
      count--;
    }
    if (v[i] > count)
      q.push(v[i]);
    else {
      ans += v[i];
      count = v[i] - 1;
    }
  }
  while (!q.empty() && count > 0) {
    ans += count;
    count--;
    q.pop();
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```