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.Collections;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

import static java.util.stream.Collectors.*;

import java.util.ArrayList;

public class D {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while (T-- != 0) {
			long n = sc.nextLong();
			Map<Long, Long> candies = new HashMap<Long, Long>();
			for (int i = 0; i < n; i++) {
				long a = sc.nextLong();
				if (candies.containsKey(a)) {
					long count = candies.get(a);
					candies.put(a, count + 1);
				} else {
					candies.put(a, (long) 1);
				}
			}
			
//			Map<Long, Long> sorted = candies
//					.entrySet()
//					.stream()
//					.sorted(Collections.reverseOrder(Map.Entry.comparingByValue()))
//					.collect(toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2,
//			                LinkedHashMap::new));
			List<Long> values = new ArrayList<>(candies.values());
			Collections.sort(values, Collections.reverseOrder());
			Iterator<Long> it = values.iterator();
			long count = 0;
			long prev = -1;
			while (it.hasNext()) {
				long value = it.next();
				if (prev == -1) {
					prev = value;
					count += prev;
				} else if (prev == 1) {
					break;
				}  else if (prev <= value) {
					prev--;
					count += prev;
				} else {
					prev = value;
					count += prev;
				}
			}
			System.out.println(count);
		}
	}

}