#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2e5 + 5;
const int inf = 0x3f3f3f3f;
struct node {
  int n, f;
  bool operator<(const node &c) const {
    if (n != c.n) {
      return n < c.n;
    } else {
      return f < c.f;
    }
  }
} a[N];
bool cmp(node &n1, node &n2) {
  if (n1.n != n2.n) {
    return n1.n > n2.n;
  } else {
    return n1.f > n2.f;
  }
}
priority_queue<node> q;
int main() {
  std::ios::sync_with_stdio(false);
  int T, n, d, f;
  cin >> T;
  while (T--) {
    while (!q.empty()) q.pop();
    cin >> n;
    for (int i = 0; i <= n + 1; i++) {
      a[i].n = a[i].f = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> d >> f;
      a[d].n++;
      if (f) a[d].f++;
    }
    for (int i = 1; i <= n; i++) {
      node tt = a[i];
      if (tt.n != 0) q.push(tt);
    }
    node t = q.top();
    int fans = t.f;
    int ans = t.n;
    int pre = t.n;
    pre--;
    q.pop();
    int len = q.size();
    for (int i = 2; pre && (!q.empty()); i++) {
      t = q.top();
      if (pre < t.n) {
        q.pop();
        t.n = pre;
        t.f = min(pre, t.f);
        q.push(t);
      } else if (t.n == pre) {
        ans += t.n;
        fans += t.f;
        pre--;
        q.pop();
      } else if (t.n < pre) {
        ans += t.n;
        pre = t.n - 1;
        fans += t.f;
        q.pop();
      }
    }
    cout << ans << ' ' << fans << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const long long mod = 1e9 + 7;
const int N = 2e5 + 5;
const int inf = 0x3f3f3f3f;
struct node {
  int n, f;
  bool operator<(const node &c) const {
    if (n != c.n) {
      return n < c.n;
    } else {
      return f < c.f;
    }
  }
} a[N];
bool cmp(node &n1, node &n2) {
  if (n1.n != n2.n) {
    return n1.n > n2.n;
  } else {
    return n1.f > n2.f;
  }
}
priority_queue<node> q;
int main() {
  std::ios::sync_with_stdio(false);
  int T, n, d, f;
  cin >> T;
  while (T--) {
    while (!q.empty()) q.pop();
    cin >> n;
    for (int i = 0; i <= n + 1; i++) {
      a[i].n = a[i].f = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> d >> f;
      a[d].n++;
      if (f) a[d].f++;
    }
    for (int i = 1; i <= n; i++) {
      node tt = a[i];
      if (tt.n != 0) q.push(tt);
    }
    node t = q.top();
    int fans = t.f;
    int ans = t.n;
    int pre = t.n;
    pre--;
    q.pop();
    int len = q.size();
    for (int i = 2; pre && (!q.empty()); i++) {
      t = q.top();
      if (pre < t.n) {
        q.pop();
        t.n = pre;
        t.f = min(pre, t.f);
        q.push(t);
      } else if (t.n == pre) {
        ans += t.n;
        fans += t.f;
        pre--;
        q.pop();
      } else if (t.n < pre) {
        ans += t.n;
        pre = t.n - 1;
        fans += t.f;
        q.pop();
      }
    }
    cout << ans << ' ' << fans << endl;
  }
  return 0;
}

```