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 class KV {
        public int k;
        public int v;
        public KV(int k, int v) {
            this.k = k;
            this.v = v;
        }
    }
    public static void main(String[] args) {
        new Main().run();
    }

    Scanner sc = new Scanner(System.in);

    void run() {
        int q = sc.nextInt();
        HashMap<Integer, KV> values = new HashMap<>();
        ArrayList<KV> list = new ArrayList<>();

        for (int i = 0; i < q; i++) {
            int n = sc.nextInt();
            values.clear();
            list.clear();

            for (int j = 0; j < n; j++) {
                int t = sc.nextInt();
                if (values.get(t) == null) {
                    values.put(t, new KV(t, 1));
                } else {
                    KV kv = values.get(t);
                    kv.v++;
                    values.put(t, kv);
                }
            }

            list.addAll(values.values());
            list.sort(new Comparator<KV>() {
                @Override
                public int compare(KV o1, KV o2) {
                    if (o1.v > o2.v) return 1;
                    else if (o1.v < o2.v) return -1;
                    return 0;
                }
            });

            int minMax = list.get(list.size() - 1).v;

            list.remove(list.get(list.size() - 1));

            int s = minMax;
            while(minMax != 1 && list.size() > 0) {
                int t = list.get(list.size() - 1).v;

                if (minMax <= t) minMax--;
                else minMax = t;
                s += minMax;

                list.remove(list.get(list.size() - 1));
            }

            System.out.println(s);
        }

    }
}