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 D_CandyBox {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    int mod = 1000000007;

    public static void main(String[] args)throws IOException{
        FastReader fr = new FastReader();
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int test = fr.nextInt();
        int n;
        int[] arr;
        int t;
        while(test-->0){
            HashMap<Integer,Integer> map = new HashMap<>();
            n = fr.nextInt();
            while(n-->0){
                t = fr.nextInt();
                if(map.containsKey(t)){
                    map.put(t,map.get(t)+1);
                }
                else{
                    map.put(t,1);
                }
            }

            Queue<Integer> q = new PriorityQueue<>(Collections.reverseOrder());
            Iterator<Integer> it = map.keySet().iterator();
            while (it.hasNext()){
                q.add(map.get(it.next()));
            }
            int ans = q.peek();
            q.poll();
            int req=ans-1;
            while(!q.isEmpty() && req!=0){
                t = q.peek();
                q.poll();
                if(t>=req){
                    ans+=req;
                    req--;
                }
                else{
                    ans+=t;
                    req=t-1;
                }
            }
            bw.write(ans+"\n");
        }
        bw.flush();
    }
}