#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 9;
int q, n;
struct T {
  int a, b;
  bool operator<(T t) const {
    if (a + b == t.a + t.b)
      return b < t.b;
    else
      return a + b < t.a + t.b;
  }
};
vector<T> a;
priority_queue<T> qu;
int main() {
  scanf("%d", &q);
  int t, mx, sum1, sum2, f;
  while (q--) {
    cin >> n;
    a.clear();
    a.resize(n + 2);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &t, &f);
      if (f == 0)
        a[t].a++;
      else
        a[t].b++;
    }
    while (!qu.empty()) qu.pop();
    for (int i = 1; i <= n; i++)
      if (a[i].a || a[i].b) qu.push(a[i]);
    T t1 = qu.top();
    qu.pop();
    sum1 = t1.a + t1.b;
    sum2 = t1.b;
    while (!qu.empty()) {
      T t2 = qu.top();
      qu.pop();
      if (t2.a + t2.b >= t1.a + t1.b) {
        int c = t2.a + t2.b - t1.a - t1.b + 1;
        if (t2.a >= c)
          qu.push((T){t2.a - c, t2.b});
        else {
          if (t2.a + t2.b > c) qu.push((T){0, t2.b - (c - t2.a)});
        }
      } else {
        t1 = t2;
        sum1 += t1.a + t1.b;
        sum2 += t2.b;
      }
    }
    printf("%d %d\n", sum1, sum2);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 + 9;
int q, n;
struct T {
  int a, b;
  bool operator<(T t) const {
    if (a + b == t.a + t.b)
      return b < t.b;
    else
      return a + b < t.a + t.b;
  }
};
vector<T> a;
priority_queue<T> qu;
int main() {
  scanf("%d", &q);
  int t, mx, sum1, sum2, f;
  while (q--) {
    cin >> n;
    a.clear();
    a.resize(n + 2);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &t, &f);
      if (f == 0)
        a[t].a++;
      else
        a[t].b++;
    }
    while (!qu.empty()) qu.pop();
    for (int i = 1; i <= n; i++)
      if (a[i].a || a[i].b) qu.push(a[i]);
    T t1 = qu.top();
    qu.pop();
    sum1 = t1.a + t1.b;
    sum2 = t1.b;
    while (!qu.empty()) {
      T t2 = qu.top();
      qu.pop();
      if (t2.a + t2.b >= t1.a + t1.b) {
        int c = t2.a + t2.b - t1.a - t1.b + 1;
        if (t2.a >= c)
          qu.push((T){t2.a - c, t2.b});
        else {
          if (t2.a + t2.b > c) qu.push((T){0, t2.b - (c - t2.a)});
        }
      } else {
        t1 = t2;
        sum1 += t1.a + t1.b;
        sum2 += t2.b;
      }
    }
    printf("%d %d\n", sum1, sum2);
  }
  return 0;
}

```