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.

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
  public static void main(String []args) {
    Scanner sc = new Scanner(System.in);
    int Q = sc.nextInt();
    for (int i = 0; i < Q; i++) {
      int n = sc.nextInt();
      int[] numPresent = new int[n+1];
      numPresent[0] = n;
      int[] types = new int[n];
      for (int j = 0; j < n; j++) {
        int type = sc.nextInt();
        types[type-1] = types[type-1] + 1;
        numPresent[types[type-1]] = numPresent[types[type-1]] + 1;
        numPresent[types[type-1]-1] = numPresent[types[type-1]-1] - 1;
      }
      
      for (int j = n; j > 0; j--) {
        if (numPresent[j] > 1) {
          int extraTypes = numPresent[j] - 1;
          numPresent[j] = 1;
          numPresent[j-1] = numPresent[j-1] + extraTypes;
        }
      }
      int maxCandy = 0;
      for (int j = 0; j <= n; j++) {
        if (numPresent[j] > 0)
          maxCandy += j;
      }
      System.out.println(maxCandy);
    }
  }
}