#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int v[MAXN];
pair<int, int> m[MAXN];
priority_queue<int> q;
int main() {
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int n;
    scanf("%d", &n);
    while (q.size()) q.pop();
    for (int i = 1; i <= n; i++) v[i] = 0;
    for (int i = 1; i <= n; i++) m[i].first = m[i].second = 0;
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      if (!v[a]) v[a] = ++cnt;
      m[v[a]].first++;
      if (b == 1) m[v[a]].second++;
    }
    sort(m + 1, m + 1 + cnt);
    int tval = m[cnt].first, ans = 0, yi = 0, conti = 0;
    m[0].first = 0;
    for (int i = cnt; i >= 0; i--) {
      if (i != cnt) tval = min(tval, m[i + 1].first);
      if (i != cnt && (m[i + 1].first != m[i].first)) {
        while (q.size() && (tval > m[i].first) && (tval > 0)) {
          ans += tval;
          yi += min(q.top(), tval);
          assert(q.size());
          q.pop();
          tval--;
        }
      }
      q.push(m[i].second);
    }
    printf("%d %d\n", ans, yi);
  }
  return 0;
}

### 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;
const int MAXN = 2e5 + 5;
int v[MAXN];
pair<int, int> m[MAXN];
priority_queue<int> q;
int main() {
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int n;
    scanf("%d", &n);
    while (q.size()) q.pop();
    for (int i = 1; i <= n; i++) v[i] = 0;
    for (int i = 1; i <= n; i++) m[i].first = m[i].second = 0;
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      if (!v[a]) v[a] = ++cnt;
      m[v[a]].first++;
      if (b == 1) m[v[a]].second++;
    }
    sort(m + 1, m + 1 + cnt);
    int tval = m[cnt].first, ans = 0, yi = 0, conti = 0;
    m[0].first = 0;
    for (int i = cnt; i >= 0; i--) {
      if (i != cnt) tval = min(tval, m[i + 1].first);
      if (i != cnt && (m[i + 1].first != m[i].first)) {
        while (q.size() && (tval > m[i].first) && (tval > 0)) {
          ans += tval;
          yi += min(q.top(), tval);
          assert(q.size());
          q.pop();
          tval--;
        }
      }
      q.push(m[i].second);
    }
    printf("%d %d\n", ans, yi);
  }
  return 0;
}

```