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.*;
import java.io.*;
import java.lang.*;
public class tyj {
    public static void main (String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t > 0) {
        	t--;
	        int n = sc.nextInt();
	        HashMap<Integer, Integer> map = new HashMap<>();
	        for(int i = 0; i < n; i++) {
	        	int x = sc.nextInt();
	        	if(map.containsKey(x)) {
	        		map.put(x, map.get(x) + 1);
	        	}
	        	else {
	        		map.put(x,  1);
	        	}
	        }
	        ArrayList<Integer> arr = new ArrayList<>();
	        for(Map.Entry<Integer, Integer> itr : map.entrySet()) {
	        	arr.add(itr.getValue());
	        }
	        Collections.sort(arr);
	        int ans = 0;
	        //for(Integer i : arr) System.out.print(i + " ");
	        //System.out.println();
	        HashMap<Integer, Integer> map1 = new HashMap<>();
	        int last = Integer.MAX_VALUE;
	        for(int i = arr.size() - 1; i >= 0; i--) {
	        	int x = arr.get(i);
	        	if(x < last) {
	        		ans += x;
	        		last = x;
	        	}
	        	else if(last - 1 > 0){
	        		ans += last - 1;
	        		last--;
	        	}
	        }
	        System.out.println(ans);
        }
        
    }
    
}

/*
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
 */