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.Arrays;
import java.util.Scanner;

public class candies 
{
	public static void main(String[] args) {
		
		Scanner in  = new Scanner(System.in);
		
		
		int iter = in.nextInt();
		
		for(int i = 0; i < iter; i++)
		{
			int q = in.nextInt();
			int arr[] = new int[q];
			for(int x = 0; x < q; x++)
			{
				arr[in.nextInt()-1] += 1; 
			}
			Arrays.sort(arr);
			
			int sum = 0;
			int biggest = Integer.MAX_VALUE;
			for(int t = q-1; t >=0; t--)
			{
				if(biggest >= arr[t])
				{
					sum += arr[t];
					biggest = arr[t] -1;
				}else 
				{
					sum += biggest;
					biggest--;
				}
				if(biggest < 0)
				{
					break;
				}
			}
			System.out.println(sum);
			
		}
		
	}
}