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.lang.*;
import java.io.*;

public class N5 {
    public static void main(String[] Args){
        Scanner scan=new Scanner(System.in);
        int t=scan.nextInt();
        while(t-->0) {
            int n = scan.nextInt();
            Map<Integer, entry> m = new HashMap<>();
            for(int i=0;i<n;i++){
                int type=scan.nextInt();
                int f=scan.nextInt();
                m.putIfAbsent(type,new entry(0,0));
                entry temp=m.get(type);
                temp.v++;
                if(f==0){
                    temp.l++;
                }
            }
            PriorityQueue<entry> q=new PriorityQueue<>(n,new myComp());
            for(Integer i: m.keySet()){
                q.add(m.get(i));
            }
            int last=q.remove().v;
            int ans=0;
            ans+=last;
            ArrayList<Integer> ps=new ArrayList<>();
            ps.add(last);
            while(!q.isEmpty()&&last>1){
                int val=q.remove().v;
                if(val>=last) {
                    last--;
                    ans+=last;
                    ps.add(last);
                }
                else{
                    ans+=val;
                    ps.add(val);
                    last=val;
                }
            }
            int ans2=0;
            for(int i:ps){
                int type=-1;
                int max=Integer.MIN_VALUE;
                for(Integer j:m.keySet()){
                    entry temp=m.get(j);
                    if(temp.v>=i){
                        int num=Math.min(temp.v-temp.l,i);
                        if(num>max){
                            max=num;
                            type=j;
                        }
                    }
                }
                m.remove(type);
                ans2+=max;
            }
            System.out.println(ans+" "+ans2);
        }
    }
    static class entry{
        int v;
        int l;
        entry(int v,int l){
            this.v=v;
            this.l=l;
        }
    }
    static class myComp implements Comparator<entry>{
        @Override
        public int compare(entry o1, entry o2) {
            if(o1.v!=o2.v){
                return -(o1.v-o2.v);
            }
            else{
                return o1.l-o2.l;
            }
        }
    }
}