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 k
{
    static int f[]=new int[200005];
    
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t=0;
        t=sc.nextInt();
        while(t>0)
        {
            int n=sc.nextInt();
            List<Integer> l1=new ArrayList<>();
            for(int i=0;i<n;i++)
            {
                int a=sc.nextInt();
                if(f[a]==0)
                    l1.add(a);
                f[a]++;
            }
            List<Integer> l=new ArrayList<>();
            for(int i=0;i<l1.size();i++)
               {
                   int val=l1.get(i);
                   l.add(f[val]);
               }
            Collections.sort(l,Collections.reverseOrder());
            int sum=0;int tt=100000000;
            for(int i=0;i<l.size();i++)
            {
                if (tt >l.get(i)) {
                    tt =l.get(i);
                } else if (tt <=l.get(i)) {
                    tt -= 1;
                } 
                if (tt <= 0)
                    break;
                sum += tt;
            }
            for (int i = 0; i < l1.size(); i++) {
				int val = l1.get(i);
				f[val] = 0;
			}
            System.out.println(sum);
            t--;
        }
    }
}