#include <bits/stdc++.h>
using namespace std;
int II() {
  int q;
  scanf("%d", &q);
  return q;
}
long long LII() {
  long long q;
  scanf("%lld", &q);
  return q;
}
struct node {
  int l, r;
  bool operator<(const node &seg) const { return r < seg.r; }
};
void fun(vector<pair<int, int>> v) {
  sort(v.begin(), v.end());
  int last = v.back().first;
  priority_queue<node> q;
  int ans = 0, ans1 = 0;
  while (last > 0) {
    while (v.size() > 0 && v.back().first >= last) {
      q.push({v.back().first, v.back().second});
      v.pop_back();
    }
    if (q.size() == 0)
      last--;
    else {
      auto it = q.top();
      q.pop();
      ans += last;
      ans1 += min(last, it.r);
      last--;
    }
  }
  cout << ans << " " << ans1 << endl;
}
void solve() {
  int N;
  cin >> N;
  unordered_map<int, int> mp, mp1;
  for (int i = 1; i <= N; i++) {
    int a, type;
    cin >> a >> type;
    mp[a]++;
    if (type == 1) mp1[a]++;
  }
  vector<pair<int, int>> v;
  for (auto it : mp) {
    v.push_back({it.second, mp1[it.first]});
  }
  fun(v);
}
int32_t main() {
  int tt;
  cin >> tt;
  int cas = 1;
  while (tt--) {
    solve();
  }
}

### Prompt

Please formulate 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;
int II() {
  int q;
  scanf("%d", &q);
  return q;
}
long long LII() {
  long long q;
  scanf("%lld", &q);
  return q;
}
struct node {
  int l, r;
  bool operator<(const node &seg) const { return r < seg.r; }
};
void fun(vector<pair<int, int>> v) {
  sort(v.begin(), v.end());
  int last = v.back().first;
  priority_queue<node> q;
  int ans = 0, ans1 = 0;
  while (last > 0) {
    while (v.size() > 0 && v.back().first >= last) {
      q.push({v.back().first, v.back().second});
      v.pop_back();
    }
    if (q.size() == 0)
      last--;
    else {
      auto it = q.top();
      q.pop();
      ans += last;
      ans1 += min(last, it.r);
      last--;
    }
  }
  cout << ans << " " << ans1 << endl;
}
void solve() {
  int N;
  cin >> N;
  unordered_map<int, int> mp, mp1;
  for (int i = 1; i <= N; i++) {
    int a, type;
    cin >> a >> type;
    mp[a]++;
    if (type == 1) mp1[a]++;
  }
  vector<pair<int, int>> v;
  for (auto it : mp) {
    v.push_back({it.second, mp1[it.first]});
  }
  fun(v);
}
int32_t main() {
  int tt;
  cin >> tt;
  int cas = 1;
  while (tt--) {
    solve();
  }
}

```