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.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author Andy Phan
 */
public class p1183d {
    public static void main(String[] args) {
        FS in = new FS(System.in);
        
        int q = in.nextInt();
        while(q-->0) {
            int n = in.nextInt();
            Integer[] num = new Integer[n];
            Arrays.fill(num, 0);
            for(int i = 0; i < n; i++) num[in.nextInt()-1]++;
            Arrays.sort(num);
            int prev = 987654321;
            int tot = 0;
            for(int i = n-1; i >= 0; i--) {
                if(prev <= 0) break;
                prev = Math.min(prev-1, num[i]);
                tot += prev;
            }
            System.out.println(tot);
        }
    }
    
    static class FS {
        BufferedReader in;
        StringTokenizer t;
        
        public FS(InputStream i) {
            in = new BufferedReader(new InputStreamReader(i));
        }
        
        public String next() {
            try {
                while(t == null || !t.hasMoreElements()) {
                    t = new StringTokenizer(in.readLine());
                }
            } catch(Exception e) {}
            return t.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}