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.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;

public class G1183 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int Q = in.nextInt();
        StringBuilder output = new StringBuilder();
        for (int q=0; q<Q; q++) {
            int N = in.nextInt();
            Candy[] candies = new Candy[N+1];
            for (int n=0; n<=N; n++) {
                candies[n] = new Candy();
            }
            for (int n=0; n<N; n++) {
                int a = in.nextInt();
                int k = in.nextInt();
                candies[a].count++;
                if (k == 0) {
                    candies[a].keepCount++;
                }
            }
            Arrays.sort(candies, new Comparator<Candy>() {
                @Override
                public int compare(Candy o1, Candy o2) {
                    return o2.count - o1.count;
                }
            });
            int total = 0;
            int totalNoKeep = 0;
            Queue<Candy> queue = new PriorityQueue<>(new Comparator<Candy>(){
                @Override
                public int compare(Candy o1, Candy o2) {
                    int noKeep1 = o1.count - o1.keepCount;
                    int noKeep2 = o2.count - o2.keepCount;
                    return noKeep2 - noKeep1;
                }
            });
            int idx = 0;
            for (int max = N; max > 0; max--) {
                while (idx < N && candies[idx].count == max) {
                    queue.add(candies[idx]);
                    idx++;
                }
                if (!queue.isEmpty()) {
                    Candy candy = queue.poll();
                    total += max;
                    totalNoKeep += Math.min(max, candy.count - candy.keepCount);
                }
            }
            output.append(total).append(' ').append(totalNoKeep).append('\n');
        }
        System.out.print(output);
    }

    static class Candy {
        int count;
        int keepCount;
    }

}