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.*;
import java.io.*;

public class candybox {
	public void run() throws Exception {
		Scanner file = new Scanner(System.in);
		int times = file.nextInt();
		for (int i = 0; i < times; i++) {
			int n = file.nextInt();
			int[] w = new int[n + 1];
			for (int j = 0; j < n; j++) w[file.nextInt()]++;
			Arrays.sort(w);
			HashSet<Integer> g = new HashSet();
			int cur = 0, prev = n + 2, count = 1, ww = w[w.length - 1];
			for (int j = w.length - 1; j >= 0; j--) {
				if (w[j] == 0 || prev <= 0) break;
				if (!g.contains(w[j])) {
					cur += w[j];
					g.add(w[j]);
					prev = w[j];
				}
				else {
					cur += prev - 1;
					g.add(prev - 1);
					prev -= 1;
				}
				//System.out.print(cur + " "); System.out.println(prev);
			}
			//System.out.println();
			System.out.println(cur);
			//System.out.println();
		}
	}

	public static void main(String[] args) throws Exception {
		new candybox().run();
	}

}