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.*;
import java.lang.*;
	

public class q4{
	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int q,n,ans = 0;
		q = in.nextInt();
		for (int i = 0;i < q ;i++ ) {
			n = in.nextInt();
			ans = 0;
			int[] a = new int[n];
			int[] b = new int[n+1];
			for (int j = 0; j<n; j++) {
				a[j] = in.nextInt();
				b[a[j]]++;
			}

			Arrays.sort(b);
			int available = b[n];
			// System.out.println(available);
			for (int j = n; j>=1; j--) {
				if (b[j] >= available) {
					ans += available;
					available = available-1;
				}
				else{
					ans+=b[j];
					available = b[j]-1;
				}
				
				if (available<=0) {
					break;
				}
				// System.out.println(b[j] + " " + available);
			}
			System.out.println(ans);
		}	
	}
}