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 Candybox
{
    public static void main (String[]args)
        {
            Scanner sc = new Scanner(System.in);
            int t =sc.nextInt();
            while (t-->0)
            {
                int n =sc.nextInt();
                int [] a = new int[n];
                HashSet<Integer> c = new HashSet<Integer>();
                for(int i =0;i<n;i++)
                {
                    a[i] = sc.nextInt();
                    c.add(a[i]);
                }
                Map<Integer, Integer> freq = new LinkedHashMap<Integer, Integer>();
                for (int i = 0; i < n; i++)
                {
                    if (freq.containsKey(a[i]) == false)
                        freq.put(a[i], 1);
                    else
                        {
                        int oldval = freq.get(a[i]);
                        int newval = oldval + 1;
                        freq.put(a[i], newval);
                    }
                }
                Set<Map.Entry<Integer, Integer>> hmap = freq.entrySet();
                int count[] = new int[c.size()];
                HashSet<Integer> check = new HashSet<Integer>();
               int k =0;
                for (Map.Entry<Integer, Integer> data : hmap)
                {
                    count[k]=data.getValue();
                    k++;
                }
                Arrays.sort(count);
                long ans = 0l;
           
                for(int i =c.size()-1;i>=0;i--)
                {
                    if(i==c.size()-1)
                    {

                        ans=count[i];
                        check.add(count[i]);

                    }
                    else if(check.contains(count[i]))
                    {
                        if(count[i+1]-1>0)
                        {

                            ans +=( count[i + 1] - 1);

                        count[i]=count[i+1]-1;
                        check.add(count[i+1]-1);
                        }
                        else
                            break;
                    }
                    else if(!check.contains(count[i]))
                    {
                        ans += count[i];
                        check.add(count[i]);

                    }
                }
                System.out.println(ans);


            }
        }
    }