#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
void solve() {
  long long n;
  cin >> n;
  map<long long, pair<long long, long long> > a;
  for (long long i = 0; i < n; i++) {
    long long x, t;
    cin >> x >> t;
    if (t == 0)
      a[x].first++;
    else
      a[x].second++;
  }
  multiset<pair<long long, long long> > help;
  for (auto i : a) {
    help.insert({-i.second.first - i.second.second, i.second.second});
  }
  multiset<long long> good;
  long long ans1 = 0;
  long long ans2 = 0;
  long long now = 1e9;
  while (!help.empty() && now > 0) {
    while (!help.empty() && (help.begin()->first) * -1 >= now) {
      auto q = *help.begin();
      good.insert(-q.second);
      help.erase(help.begin());
    }
    if (!good.empty()) {
      long long x = *good.begin();
      good.erase(good.begin());
      x = -x;
      ans1 += now;
      ans2 += min(now, x);
      now--;
    } else {
      auto q = *help.begin();
      now = -q.first;
    }
  }
  while (!good.empty() && now > 0) {
    long long x = *good.begin();
    good.erase(good.begin());
    x = -x;
    ans1 += now;
    ans2 += min(now, x);
    now--;
  }
  cout << ans1 << " " << ans2 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
void solve() {
  long long n;
  cin >> n;
  map<long long, pair<long long, long long> > a;
  for (long long i = 0; i < n; i++) {
    long long x, t;
    cin >> x >> t;
    if (t == 0)
      a[x].first++;
    else
      a[x].second++;
  }
  multiset<pair<long long, long long> > help;
  for (auto i : a) {
    help.insert({-i.second.first - i.second.second, i.second.second});
  }
  multiset<long long> good;
  long long ans1 = 0;
  long long ans2 = 0;
  long long now = 1e9;
  while (!help.empty() && now > 0) {
    while (!help.empty() && (help.begin()->first) * -1 >= now) {
      auto q = *help.begin();
      good.insert(-q.second);
      help.erase(help.begin());
    }
    if (!good.empty()) {
      long long x = *good.begin();
      good.erase(good.begin());
      x = -x;
      ans1 += now;
      ans2 += min(now, x);
      now--;
    } else {
      auto q = *help.begin();
      now = -q.first;
    }
  }
  while (!good.empty() && now > 0) {
    long long x = *good.begin();
    good.erase(good.begin());
    x = -x;
    ans1 += now;
    ans2 += min(now, x);
    now--;
  }
  cout << ans1 << " " << ans2 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```