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 Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		int q=in.nextInt();
		for(int i=0;i<q;i++) {
			work();
		}
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return b==0?a:gcd(b,a%b);
	}
	
	void work() {
		int n=in.nextInt();
		int[] count1=new int[n+1];
		int[] count2=new int[n+1];
		for(int i=0;i<n;i++) {
			int a=in.nextInt();
			int f=in.nextInt();
			count1[a]++;
			if(f==1)count2[a]++;
		}
		ArrayList<int[]> list=new ArrayList<>();
		for(int i=1;i<=n;i++) {
			if(count1[i]>0) list.add(new int[] {count1[i],count2[i]});
		}
		Collections.sort(list,new Comparator<int[]>() {
			public int compare(int[] arr1,int[] arr2) {
				return arr2[0]-arr1[0];
			}
		});
		PriorityQueue<Integer> pq=new PriorityQueue<>();
		int ret1=0;
		int ret2=0;
		for(int i=0,cur=1;cur>0;cur--) {
			if(pq.size()==0&&i==list.size()) break;
			if(pq.size()==0) {
				cur=list.get(i)[0];
			}
			while(i<list.size()&&list.get(i)[0]==cur) {
				pq.add(-list.get(i)[1]);
				i++;
			}
			ret1+=cur;
			ret2+=Math.min(cur,-pq.poll());
		}
		out.println(ret1+" "+ret2);
	}
}



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}