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 Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int T = scan.nextInt();
        for (int A=0;A<T;A++){
            int n = scan.nextInt();
            HashMap<Integer,Integer> map = new HashMap<>();
            ArrayList<Integer> arr = new ArrayList<>();
            int count = 0;
            for (int i=0;i<n;i++) {
            	int s = scan.nextInt();
            	if (map.containsKey(s)) {
            		int a = map.get(s);
            		arr.set(a,arr.get(a)+1);
            	}
            	else {
            		map.put(s,count++);
            		arr.add(1);
            	}
            }
            Collections.sort(arr,new Comp());
//            System.out.println(arr);
            count = arr.get(0);
            int min  = arr.get(0);
            for (int i=1;i<arr.size();i++) {
            	if (arr.get(i)<min) {
            		count += arr.get(i);
            		min = arr.get(i);
            	}
            	else {
            		if (min-1>=0) {
            			count += min-1;
            			min = min-1;
            		}
            		else {
            			break;
            		}
            	}
            }
            System.out.println(count);
        }
    }
}
class Comp implements Comparator<Integer>{
    public int compare(Integer a,Integer b){
        return (int)b-(int)a;
    }
}