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 javax.lang.model.type.PrimitiveType;
import java.math.BigInteger;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        TaskA task = new TaskA();
        task.solve();
        System.out.flush();
    }

    static class TaskA {
        Scanner scanner;
        final int INF = (int) 1e9 + 7;
        int n;

        public void solve() {
            scanner = new Scanner(System.in);
            int Q = scanner.nextInt();
            Map<Integer, Integer> map;
            for(int q = 0; q < Q; ++ q) {
                int n = scanner.nextInt();
                map = new TreeMap<>();
                for(int i = 0; i < n; ++ i) {
                    int a = scanner.nextInt();
                    map.put(a, (map.containsKey(a) ? map.get(a) + 1 : 1));
                }
                List<Integer> list = new ArrayList<>();
                for(int value : map.values())
                    list.add(value);
                Collections.sort(list, Collections.reverseOrder());
                int tec = list.get(0), ans = tec;
                for(int i = 1; i < list.size(); ++ i) {
                    tec = Math.min(tec - 1, list.get(i));
                    if(tec < 0) break;
                    ans += tec;
                }
                System.out.println(ans);
            }
        }
    }
}