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.io.*;
import java.util.*;
public class Main {
	static BufferedReader br;
	static PrintWriter pr;
	static int cin() throws Exception
	{
		return Integer.valueOf(br.readLine());
	}
	static int[] split() throws Exception
	{
		String[] cmd=br.readLine().split(" ");
		int[] ans=new int[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Integer.valueOf(cmd[i]);
		}
		return ans;
	}
	static long[] splitL() throws IOException
	{
		String[] cmd=br.readLine().split(" ");
		long[] ans=new long[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Long.valueOf(cmd[i]);
		}
		return ans;
	}
	static int sum(long n)
	{
		int ans=0;
		while(n!=0)
		{
			ans=ans+(int)(n%10);
			n=n/10;
		}
		return ans;
	}
	static long gcd(long x,long y)
	{
		if(x==0)
			return y;
		if(y==0)
			return x;
		if(x>y)
			return gcd(x%y,y);
		else
			return gcd(x,y%x);
	}
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		br=new BufferedReader(new InputStreamReader(System.in));
		pr=new PrintWriter(new OutputStreamWriter(System.out));
		int cases=cin();
		while(cases!=0)
		{
			cases--;
			int n=cin();
			int[]arr=split();
			int[]cnt=new int[n+1];
			for(int i=0;i<n;i++)
			{
				cnt[arr[i]]++;
			}
			ArrayList<Integer>list=new ArrayList<>();
			for(int i=0;i<=n;i++)
			{
				if(cnt[i]!=0)
					list.add(cnt[i]);
			}
			Collections.sort(list,Collections.reverseOrder());
			long ans=0;
			int mx=list.get(0);
			for(int i=0;i<list.size();i++)
			{
				if(list.get(i)>=mx)
				{
					ans=ans+(long)mx;
					mx--;
				}
				else
				{
					mx=list.get(i);
					ans=ans+(long)mx;
					mx--;
				}
				if(mx<=0)
					break;
			}
			System.out.println(ans);
		}
	}
}