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 candybox {
    public static void main(String[] args) throws Exception{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int q = Integer.parseInt(in.readLine());
        for(int i = 0; i < q; i++){
            int n = Integer.parseInt(in.readLine());
            int[] candies = new int[n];
            StringTokenizer st = new StringTokenizer(in.readLine());
            for(int j = 0; j < n; j++){
                candies[j] = Integer.parseInt(st.nextToken());
            }
            Map<Integer, Integer> map = new TreeMap<>();
            for(int j: candies){
                if(!map.containsKey(j)){
                    map.put(j, 1);
                }
                else{
                    map.put(j, map.get(j) + 1);
                }
            }
            Set<Integer> already = new HashSet<>();
            int ans = 0;
            for(int key: map.keySet()){
                int max = -1;
                for(int j = 1; j <= map.get(key); j++){
                    if(already.contains(j)){
                        continue;
                    }
                    max = Math.max(max, j);
                }
                if(max != -1){
                    already.add(max);
                    ans += max;
                }
            }
            System.out.println(ans);
        }
    }
}