1183_D. Candy Box (easy version)_17102
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.
import java.io.*;
import java.util.*;

public class Cony {



	public static void main(String[] args) throws IOException
	{

		BufferedReader jk = new BufferedReader(new InputStreamReader(System.in)) ;
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)) ; 
		int q = Integer.parseInt(jk.readLine()) ; 
		for(int j=0 ; j< q ;j++)
		{
			int n = Integer.parseInt(jk.readLine()) ;
			StringTokenizer ana = new StringTokenizer(jk.readLine()) ; 
			int tab[]= new int[n]; 
			ArrayList<Integer>s = new ArrayList<Integer>() ; 
			for(int i=0 ; i< n ;i++)
			{
				int d= Integer.parseInt(ana.nextToken())-1; 
				tab[d]++ ; 
			}
			for(int i=0 ; i<n ; i++)
			{
				s.add(tab[i]) ; 
				
			}
			Collections.sort(s);
			int res = s.get(n-1) ;
			int pre = s.get(n-1) ; 
			
			for(int i=n-2 ;i>=0 ;i-- )
			{
				int cur= s.get(i) ; 
				if(cur>=pre)
				{
					res+= Math.max(pre-1, 0) ; 
					pre =Math.max(pre-1, 0) ;
					if(pre==0) break ; 
				}
				else
				{
					pre = cur ;
					res+= cur ;
				}
			}
			out.println(res);
			
		}




		out.flush();

	}
}

