#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int N = 200200;
int a[N];
int f[N];
pair<int, int> Cnt[N];
void TestCase() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", a + i, f + i);
    a[i]--;
    Cnt[a[i]].first++;
    if (f[i]) Cnt[a[i]].second++;
  }
  sort(Cnt, Cnt + n, greater<pair<int, int> >());
  pair<int, int> Ans = pair<int, int>(0, 0);
  int Can = n;
  int Last = 0;
  priority_queue<pair<int, int> > q;
  do {
    if (!q.empty()) {
      pair<int, int> p = q.top();
      q.pop();
      int first = max(0, min(p.second, Can));
      int second = min(first, p.first);
      Ans.first += first;
      Ans.second += second;
      Can = max(0, first - 1);
    }
    while (Last < n && (q.empty() || Cnt[Last].first >= Can)) {
      q.push(pair<int, int>(Cnt[Last].second, Cnt[Last].first));
      Last++;
    }
  } while (!q.empty());
  printf("%d %d\n", Ans.first, Ans.second);
  for (int i = 0; i < n; i++) Cnt[i] = pair<int, int>(0, 0);
}
int main() {
  int T;
  cin >> T;
  while (T--) TestCase();
}

### 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 namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int N = 200200;
int a[N];
int f[N];
pair<int, int> Cnt[N];
void TestCase() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", a + i, f + i);
    a[i]--;
    Cnt[a[i]].first++;
    if (f[i]) Cnt[a[i]].second++;
  }
  sort(Cnt, Cnt + n, greater<pair<int, int> >());
  pair<int, int> Ans = pair<int, int>(0, 0);
  int Can = n;
  int Last = 0;
  priority_queue<pair<int, int> > q;
  do {
    if (!q.empty()) {
      pair<int, int> p = q.top();
      q.pop();
      int first = max(0, min(p.second, Can));
      int second = min(first, p.first);
      Ans.first += first;
      Ans.second += second;
      Can = max(0, first - 1);
    }
    while (Last < n && (q.empty() || Cnt[Last].first >= Can)) {
      q.push(pair<int, int>(Cnt[Last].second, Cnt[Last].first));
      Last++;
    }
  } while (!q.empty());
  printf("%d %d\n", Ans.first, Ans.second);
  for (int i = 0; i < n; i++) Cnt[i] = pair<int, int>(0, 0);
}
int main() {
  int T;
  cin >> T;
  while (T--) TestCase();
}

```