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 stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        for(int i = 0; i < n; i++)
        {
        	test(stdin);
        }
        //test(stdin);
        stdin.close();
    }
    public static void test(Scanner stdin)
    {
    	int n = stdin.nextInt();
    	int[] arr = new int[n];
    	for(int i = 0; i < n; i++)
    	{
    		arr[stdin.nextInt() - 1]++;
    	}
    	mix(arr);
    	Arrays.sort(arr);
    	int sum = 0;
    	int last = Integer.MAX_VALUE;
    	for(int i = n - 1; i >= 0; i--)
    	{
    		if(arr[i] >= last)
    		{
    			arr[i] = last - 1;
    		}
    		if(arr[i] <= 0)
    		{
    			break;
    		}
    		sum += arr[i];
    		last = arr[i];
    	}
    	System.out.println(sum);
    }
	public static void mix(int input[])
	{
		Random rand = new Random();
		for(int i = 0; i < input.length; i++)
		{
			int index = rand.nextInt(input.length);
			int temp = input[index];
			input[index] = input[i];
			input[i] = temp;
		}
	}
}