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.ArrayList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int tests = in.nextInt();

        for (int i = 0; i < tests; i += 1) {
            int n = in.nextInt();

            Candies[] candies = new Candies[n];
            for (int j = 0; j < n; j += 1) {
                candies[j] = new Candies();
            }
            for (int j = 0; j < n; j += 1) {
                int type = in.nextInt();
                int wanted = in.nextInt();
                candies[type - 1].add(wanted);
            }

            List<List<Candies>> byTotal = new ArrayList<List<Candies>>();
            for (int j = 0; j <= n; j += 1) {
                byTotal.add(new ArrayList<Candies>());
            }
            for (Candies c : candies) {
                byTotal.get(c.total()).add(c);
            }

            PriorityQueue<Candies> heap = new PriorityQueue<Candies>();
            int giftTotal = 0;
            int giftCost = 0;

            for (int j = n; j > 0; j -= 1) {
                for (Candies c : byTotal.get(j)) {
                    heap.add(c);
                }
                if (heap.size() > 0) {
                    giftTotal += j;
                    giftCost += heap.peek().profit(j);
                    heap.poll();
                }
            }
            System.out.println(giftTotal + " " + giftCost);
        }
    }
}

class Candies implements Comparable<Candies> {
    private int countLiked;
    private int countDisliked;

    public void add(int wanted) {
        if (wanted == 1) {
            countDisliked += 1;
        } else {
            countLiked += 1;
        }
    }

    public int total() {
        return countLiked + countDisliked;
    }

    public int profit(int taken) {
        return Math.min(countDisliked, taken);
    }

    @Override
    public int compareTo(Candies other) {
        if (countDisliked == other.countDisliked) {
            return other.countLiked - countLiked;
        }
        return other.countDisliked - countDisliked;
    }
}