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:
//   package cp;
import java.io.*;
import java.math.*;
import java.util.*;
public class Cf_one {
	long gcd(long a,long b) {if(b==0)return a;else return gcd(b,a%b);}
	void swap(long a,long b) {long temp=a;a=b;b=temp;}
	public static void main(String[] args) throws IOException {	
		PrintWriter out = new PrintWriter(System.out);
		Readers.init(System.in);
		int q=Readers.nextInt();
		for (int i = 0; i < q; i++) {
			int n=Readers.nextInt();
			int[] a=new int[n];ArrayList<Integer> arr=new ArrayList<>();
			HashMap<Integer,Integer> map=new HashMap<>();
			for (int j = 0; j < n; j++) {
				a[j]=Readers.nextInt();
				if(map.get(a[j])==null)map.put(a[j],1);
				else map.put(a[j],map.get(a[j])+1);
			}
			for(Integer temp:map.keySet()) {
				arr.add(map.get(temp));
			}
			Collections.sort(arr);Collections.reverse(arr);
			long summ=arr.get(0);
//			System.out.println(arr);
			for (int j = 1; j < arr.size();j++) {
				long temp=Math.min(arr.get(j-1)-1, arr.get(j));
				if(temp==0)break;
				summ+=temp;
				arr.set(j,(int) temp);
			}
			System.out.println(summ);
		}
		out.flush();
	}
	
}
class Readers {
 
    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 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 double nextDouble() throws IOException {
 
        return Double.parseDouble( next() );
 
    }
    
    
    static long nextLong() throws IOException{
    	
		return Long.parseLong(next());
	}
 
}