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 d2 {
	
	public static void main(String[] args) throws Exception {
	
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		int numQ = Integer.parseInt(stdin.readLine());
		
		for (int loop=0; loop<numQ; loop++) {
		
			int n = Integer.parseInt(stdin.readLine());
			int[] freq = new int[n+1];
			StringTokenizer tok = new StringTokenizer(stdin.readLine());
			for (int i=0; i<n; i++)
				freq[Integer.parseInt(tok.nextToken())]++;
			boolean[] used = new boolean[n+1];
			ArrayList<Integer> nums = new ArrayList<Integer>();
			for (int i=0; i<=n; i++) 
				if (freq[i] > 0)
					nums.add(freq[i]);
			Collections.sort(nums);
			Collections.reverse(nums);
			int ans = nums.get(0);
			int cur = ans;
			
			for (int i=1; i<nums.size(); i++) {
				if (nums.get(i) >= cur) {
					ans += cur-1;
					cur = cur-1;
				}
				else {
					ans += nums.get(i);
					cur = nums.get(i);
				}
				if (cur <= 0) break;
			}
			System.out.println(ans);
		
		}
	}
}