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 CandyBox
{
	public static void main(String args[])throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine().trim());
		while(t-- > 0)
		{
			int N = Integer.parseInt(br.readLine().trim());
			String S[] = br.readLine().split(" ");
			LinkedHashMap<Integer , Integer> map = new LinkedHashMap<Integer , Integer>();
			int ar[] = new int[N];
			int cnt = 0;
			for(int i = 0 ; i < N ; i++)
			{
				int nn = Integer.parseInt(S[i]);
				if(!map.containsKey(nn))
				{
					cnt += 1;
					map.put(nn , 1);
				}
				else
				{
					map.put(nn , 1 + map.get(nn));
				}
			}
			int A[] = new int[cnt];
			int index = 0;
			//LinkedHashSet<Integer> set = new LinkedHashSet<Integer>();
			for(Map.Entry<Integer , Integer> e : map.entrySet())
			{
				A[index++] = e.getValue();
	//			set.add(e.getValue());
			}
		/*	long sum = 0;
			for(Integer ii : set)
			{
				sum += ii;
			}*/
			Arrays.sort(A);
		//	for(int i = 0 ; i < A.length ; i++) System.out.println(A[i]);
			long sum = 0;
			int prev = -1;
			for(int i = A.length - 1 ; i >= 0 ; i--)
			{
				if(i == (A.length - 1))
				{
					sum += A[i];
					prev = A[i];
					continue;
				}
				if(A[i] >= prev)
				{
					sum += (prev - 1);
					prev -= 1;
				}
				else
				{
					sum += A[i];
					prev = A[i];
				}
				if(prev == 0)
				{
					break;
				}
			}
			pw.println(sum);
		}
		pw.flush();
	}
}