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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class candyBoxEasy {
    static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter p = new PrintWriter(System.out);

    public static void main(String arg[]) throws IOException {
        StringTokenizer st = new StringTokenizer(b.readLine());
        int x = Integer.parseInt(st.nextToken());
        while(x > 0){
            x--;
            st = new StringTokenizer(b.readLine());
            int i = Integer.parseInt(st.nextToken());
            st = new StringTokenizer(b.readLine());
            Map<Integer, Integer> ar = new HashMap<>();
            for(int y = 0; y < i; y++){
                int k = Integer.parseInt(st.nextToken());
                if(ar.keySet().contains(k)){
                    ar.put(k, ar.get(k) + 1);
                } else {
                    ar.put(k, 1);
                }
            }
            PriorityQueue<cbe> pq = new PriorityQueue<>();
            Iterator<Integer> iter = ar.keySet().iterator();
            while(iter.hasNext()){
                int k = iter.next();
                pq.add(new cbe(k, ar.get(k)));
            }
            int count = 0;
            for(int y = i; y > 0; y--){
                if(pq.isEmpty()){
                    break;
                }
                if(pq.peek().v >= y){
                    count += y;
                    pq.remove();
                }
            }
            p.println(count);
        }
        p.close();
    }
}
class cbe implements Comparable<cbe>{
    int t;
    int v;
    public cbe(int ty, int val){
        t = ty;
        v = val;
    }
    public int compareTo(cbe c){
        if(v > c.v){
            return -1;
        }
        return 1;
    }
}