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.util.HashMap;
import java.util.Map;
import java.util.Scanner;


public class CandyBox {
      public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
    
         int t = input.nextInt();
          for (int i = 0; i < t; i++) {
              Map<Integer, Integer> m1 = new HashMap<Integer, Integer>();
              Map<Integer, Integer> m2 = new HashMap<Integer, Integer>();
              
              int n = input.nextInt();
              for (int j = 0; j < n; j++) {
                  int x = input.nextInt();
                  if(m1.containsKey(x))
                      m1.put(x, m1.get(x)+1);
                  else
                      m1.put(x, 1);
                  
              }
              
              for (Integer xx : m1.keySet()) {
                  
                  if(!m2.containsKey(m1.get(xx)))
                      m2.put(m1.get(xx), 1);
                  else{
                      int y = m1.get(xx);
                     while(y>0){
                         if(m2.containsKey(y))
                         y--;
                         else
                             break;
                     }
                     m2.put(y, 1);
                  }
              }
              int count=0;
              for (Integer aa : m2.keySet()) {
                  //System.out.println(aa+" "+m2.get(aa));
                  count+=aa;
              }
              System.out.println(count);
          }
}
}