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 Solution {
	
   public static void main(String[] args) {
	   Scanner ip=new Scanner(System.in);
	   int q=ip.nextInt();
	   while(q>0) {
		   int n=ip.nextInt();
		   int arr[]=new int[n];
		   ArrayList<Integer> candy=new ArrayList<>();
		   HashSet<Integer> set=new HashSet<>();
		   for(int i=0;i<n;i++) {
			   arr[i]=ip.nextInt();
			   set.add(arr[i]);
		   }
		   candy.addAll(set);
		   HashMap<Integer,Integer> hm=new HashMap<>();
		   for(int i=0;i<n;i++) {
			   if(hm.containsKey(arr[i])) {
				   hm.put(arr[i],hm.get(arr[i])+1);
			   }else {
				   hm.put(arr[i],1);
			   }
		   }
		   HashSet<Integer> hs=new HashSet<>();
           Integer sum=0;
		   for(int i=0;i<candy.size();i++) {
			   Integer value=hm.get(candy.get(i));
			   if(!hs.contains(value)) {
				   hs.add(value);
				   sum+=value;
			   }else {
				  while(hs.contains(value) && value>0) {
					  value--;
				  }
				  hs.add(value);
				  sum+=value;
			   }
		   }
		   System.out.println(sum);
		   
		   q--;
	   } 
   }
}