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 CodeForces {
	static int z =0;
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int test = input.nextInt();
		while(test-->0){
			int n = input.nextInt();
			z=0;
			int arr[] = new int[n+1];
			for(int i =0;i<n;i++){
				arr[input.nextInt()]++;
			}
			sort(arr);
			long sum =0;
			HashSet<Integer>set = new HashSet<>();
			int target =0;
			for(int i =z-1;i>=0;i--){
				if(!set.contains(arr[i])){
					sum+=arr[i];
					set.add(arr[i]);
					target =  arr[i];
				}else{
					target--;
					if(target>0){
					sum+=target;
					set.add(target);
					}
				}
			}
			System.out.println(sum);
		}
	}
	public static void sort(int arr[]){
		ArrayList<Integer>list = new ArrayList<>();
		for(int it:arr){if(it != 0)list.add(it);}
		Collections.sort(list);
		for(int i =0;i<list.size();i++)arr[z++] = list.get(i);
	}
		

}

/*
1.make sure to include continue or return if u are handling a specific case separately!!!!
and check the constraints of every alphabet(missed to check for case 0)!!!!!!!!!

2.when u have transitions between states and its not dp try dfs (also think of topo sort)
 */