#include <bits/stdc++.h>
int asc(const void* a, const void* b) { return *(int*)b - *(int*)a; }
int max(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
int main() {
  int q;
  scanf("%d", &q);
  int ans[200005];
  int i, j;
  int n, m;
  int a[200005], b[200005];
  int x;
  for (i = 0; i < q; i++) {
    scanf("%d", &n);
    for (j = 0; j < n; j++) scanf("%d", &a[j]);
    if (n == 1)
      ans[i] = 1;
    else {
      qsort(a, n, sizeof(int), asc);
      x = 1;
      j = 1;
      m = 0;
      while (j < n) {
        while (a[j - 1] == a[j] && j < n) {
          j++;
          x++;
        }
        j++;
        b[m] = x;
        x = 1;
        m++;
        if (j == n) {
          b[m] = x;
          m++;
        }
      }
      qsort(b, m, sizeof(int), asc);
      ans[i] = b[0];
      for (j = 1; j < m; j++) {
        if (b[j] < b[j - 1])
          ans[i] += b[j];
        else {
          ans[i] += max(0, b[j - 1] - 1);
          b[j] = max(0, b[j - 1] - 1);
        }
      }
    }
  }
  for (i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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>
int asc(const void* a, const void* b) { return *(int*)b - *(int*)a; }
int max(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
int main() {
  int q;
  scanf("%d", &q);
  int ans[200005];
  int i, j;
  int n, m;
  int a[200005], b[200005];
  int x;
  for (i = 0; i < q; i++) {
    scanf("%d", &n);
    for (j = 0; j < n; j++) scanf("%d", &a[j]);
    if (n == 1)
      ans[i] = 1;
    else {
      qsort(a, n, sizeof(int), asc);
      x = 1;
      j = 1;
      m = 0;
      while (j < n) {
        while (a[j - 1] == a[j] && j < n) {
          j++;
          x++;
        }
        j++;
        b[m] = x;
        x = 1;
        m++;
        if (j == n) {
          b[m] = x;
          m++;
        }
      }
      qsort(b, m, sizeof(int), asc);
      ans[i] = b[0];
      for (j = 1; j < m; j++) {
        if (b[j] < b[j - 1])
          ans[i] += b[j];
        else {
          ans[i] += max(0, b[j - 1] - 1);
          b[j] = max(0, b[j - 1] - 1);
        }
      }
    }
  }
  for (i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}

```