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.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class q5 {
	
	static boolean check(int a) {
		String s=Integer.toString(a);
		int sum=0;
		for(int i=0;i<s.length();i++) {
			sum+=(s.charAt(i)-'0');
		}
		return (sum%4==0);
	}
	
	public static void main(String[] args) throws IOException {
		
		Reader.init(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int q=Reader.nextInt();
		while(q>0) {
			q--;
			int n=Reader.nextInt();
			int[] count=new int[n+1];
			for(int i=0;i<n;i++) count[Reader.nextInt()]++;
			ArrayList<Integer> arr=new ArrayList<Integer>();
			int lastinclude=n+1;
			for(int i=1;i<=n;i++) {
				if(count[i]>0) arr.add(count[i]);
			}
			Collections.sort(arr);Collections.reverse(arr);
			int ans=0;
			for(int i=0;i<arr.size();i++) {
				int a=arr.get(i);
				if(lastinclude<=0) break;
				if(a>=lastinclude ) {
					ans+=lastinclude-1;
					lastinclude--;
				}
				else {
					ans+=a;
					lastinclude=a;
				}
				
			}
			out.println(ans);
		}
		
		out.flush();
		
	}
}








class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String nextLine() throws IOException{
    	return reader.readLine();
    }
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }
    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}