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 GFG 
{
	public static void main (String[] args) 
	{
	    Scanner sc=new Scanner(System.in);
	    int T=sc.nextInt();
	    while(T-->0)
	    {
    	    int N=sc.nextInt();
    	    int ar[]=new int[N+1];
    	    for(int i=0;i<N;i++)
    	    {
    	        int h=sc.nextInt();
    	        ar[h]++;
    	    }
    	    long ans=0;
    	    Arrays.sort(ar);
    	    int t=N;
    	    int y=1000000;
    	    while(t>=0)
    	    {
    	        if(ar[t]==0||y==1)
    	            break;
    	        if(ar[t]<y)
    	        {
    	            ans+=ar[t];
    	            y=ar[t];
    	            t--;
    	        }
    	        else
    	        {
    	            ans+=y-1;
    	            y=y-1;
    	            t--;
    	        }
    	    }
    	    System.out.println(ans);
	    }
	}
}