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.*;
import java.io.*;
public class TestClass{
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int q,n;
		q = sc.nextInt();
		while(q-->0)
		{
			n = sc.nextInt();
			HashMap<Integer,Integer> map = new HashMap<>(n);
			for(int i = 0;i < n; i++)
			{
				int num = sc.nextInt();
				map.put(num,map.getOrDefault(num,0) + 1);
			}
			ArrayList<Integer> counts = new ArrayList<Integer>(map.values());
			Collections.sort(counts,Collections.reverseOrder());
			int giftSize = counts.get(0);
			int nxt = counts.get(0) - 1;
			for(int i=1;i<counts.size();i++)
			{
				if(nxt<=0)
					break;
				if(counts.get(i)>=nxt)
				{
					giftSize+=nxt;
					nxt-=1;
				}
				else
				{
					giftSize+=counts.get(i);
					nxt = counts.get(i)-1;
				}
			}
			System.out.println(giftSize);
		}		
	}
}