#include <bits/stdc++.h>
using ll = long long;
int solve();
int main(int argc, char* argv[]) {
  int Q = 1;
  (void)scanf("%d", &Q);
  while (Q--) solve();
  return 0;
}
static int readInt() {
  int x = 0;
  if (1 == scanf("%d", &x)) return x;
  return 0;
}
static ll readLL() {
  ll x = 0;
  if (1 == scanf("%lld", &x)) return x;
  return 0;
}
static const int N = (1 << 18);
int solve() {
  const int n = readInt();
  std::vector<int> cnt(n + 1), cnt_f(n + 1);
  for (int i = 0; i < n; ++i) {
    int a = readInt();
    int f = readInt();
    cnt[a]++;
    cnt_f[a] += f;
  }
  std::vector<std::vector<int> > tp(n + 1);
  for (int i = 1; i <= n; ++i) {
    tp[cnt[i]].push_back(cnt_f[i]);
  }
  std::pair<long long, long long> ans = {0, 0};
  std::set<std::pair<int, int>, std::greater<std::pair<int, int> > > ss;
  int gen = 0;
  for (int i = n; i > 0; --i) {
    for (int fn : tp[i]) ss.insert(std::make_pair(fn, ++gen));
    if (!ss.empty()) {
      auto it = ss.begin();
      std::pair<int, int> greater = *it;
      ss.erase(it);
      int z = greater.first;
      ans.first += i;
      ans.second += std::min(i, z);
    }
  }
  printf("%lld %lld\n", ans.first, ans.second);
  return 0;
}

### Prompt

Your task is to create a cpp solution 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 ll = long long;
int solve();
int main(int argc, char* argv[]) {
  int Q = 1;
  (void)scanf("%d", &Q);
  while (Q--) solve();
  return 0;
}
static int readInt() {
  int x = 0;
  if (1 == scanf("%d", &x)) return x;
  return 0;
}
static ll readLL() {
  ll x = 0;
  if (1 == scanf("%lld", &x)) return x;
  return 0;
}
static const int N = (1 << 18);
int solve() {
  const int n = readInt();
  std::vector<int> cnt(n + 1), cnt_f(n + 1);
  for (int i = 0; i < n; ++i) {
    int a = readInt();
    int f = readInt();
    cnt[a]++;
    cnt_f[a] += f;
  }
  std::vector<std::vector<int> > tp(n + 1);
  for (int i = 1; i <= n; ++i) {
    tp[cnt[i]].push_back(cnt_f[i]);
  }
  std::pair<long long, long long> ans = {0, 0};
  std::set<std::pair<int, int>, std::greater<std::pair<int, int> > > ss;
  int gen = 0;
  for (int i = n; i > 0; --i) {
    for (int fn : tp[i]) ss.insert(std::make_pair(fn, ++gen));
    if (!ss.empty()) {
      auto it = ss.begin();
      std::pair<int, int> greater = *it;
      ss.erase(it);
      int z = greater.first;
      ans.first += i;
      ans.second += std::min(i, z);
    }
  }
  printf("%lld %lld\n", ans.first, ans.second);
  return 0;
}

```