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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class pD {
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int q = Integer.parseInt(br.readLine());
		PrintWriter out = new PrintWriter(System.out);
		while(q-->0) {
			int n = Integer.parseInt(br.readLine());
			StringTokenizer st = new StringTokenizer(br.readLine());
			HashMap<Integer, Integer> hm = new HashMap<>();
			for (int i=0;i<n;i++) {
				int x = Integer.parseInt(st.nextToken());
				if (hm.containsKey(x)) {
					hm.put(x, hm.get(x)+1);
				}
				else {
					hm.put(x,1);
				}
			}
			PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
			for (int x: hm.values()) {
				pq.add(x);
			}
			int i = pq.poll();
			int res = i;
			i--;
			while(!pq.isEmpty()&&i>0) {
				int cur = pq.poll();
				if (cur>=i) {
					res+=i;
					i--;
				}
				else {
					res +=cur;
					i=cur-1;
				}
			}
			
			out.println(res);
		}
		br.close();
		out.close();
	}
}