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 final class D1183 {
	
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringBuilder str = new StringBuilder();
				int t = Integer.parseInt(br.readLine());
		while(t-- > 0) {
			int n = Integer.parseInt(br.readLine());
			int[]a = new int[n];
			int [] temp = new int[n];
			String[] split = br.readLine().split(" ");
			for(int i = 0;i < n; i++) {
				a[i] = Integer.parseInt(split[i]);
				temp[a[i] - 1]++;
			}
			Arrays.sort(temp);
			int ans = 0;
			ans += temp[temp.length - 1];
			int max = ans;
			for(int i = temp.length - 2;i >= 0&&  temp[i] != 0; i--) {
				
					if(temp[i] < max) {
						ans += temp[i];
						max = temp[i];
					}
					else {
						max--;
						ans += max;
					}
					if(max == 0) {
						break;
					}
				
				
			}
			str.append(ans + "\n");
			
			
		}
		System.out.print(str);
		
	}
}