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.*;

public class codeforces {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int q = sc.nextInt();
		for(int i=0;i<q;i++) {
			int n = sc.nextInt();
			TreeMap<Integer,Stack<Integer>> tr = new TreeMap<>();
			int[] countarr = new int[n+1];
			for(int j=0;j<n;j++) {
				countarr[sc.nextInt()]++;
			}
//			for(int l = 1;l<n+1;l++)
//				System.out.print(countarr[l] + " ");
//			System.out.println();
			for(int j=1;j<n+1;j++) {
				if(countarr[j]!=0) {
				if(!tr.containsKey(countarr[j])) {
					Stack<Integer> temp= new Stack<>();
					temp.push(j);
					tr.put(countarr[j], temp);
				}else {
					tr.get(countarr[j]).push(j);
				}
				}
			}
			//System.out.println(tr);
			
			int c =0;
			int last = 0;
			boolean[] v = new boolean[n+1];
			for(int k=1;!tr.isEmpty();) {
				for(int j= tr.firstEntry().getKey();!tr.firstEntry().getValue().isEmpty() && j>0;j--) {
					if(v[j]==false) {
					int te =tr.firstEntry().getValue().pop();
					//System.out.print(j + " ");
					v[j] = true;
					c+=j;
					}
				}
				last = tr.firstEntry().getKey();
				tr.pollFirstEntry();
				
			}
			//System.out.println();
			System.out.println(c);
		}
	}
}