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:
//Code by Sounak, IIEST
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.Arrays;

public class Test2{
    public static void main(String args[])throws IOException{
        FastReader sc = new FastReader();  
        StringBuilder sb = new StringBuilder();
        int test=sc.nextInt();
        while(--test>=0)
        {
            int n=sc.nextInt();
            int a[]=new int[n];
            int c[]=new int[n];
            int i,prev=0,count=0;
            for(i=0;i<n;i++)
            {
                int temp=sc.nextInt();
                c[temp-1]++;
            }
            Arrays.sort(c);
            //for(i=0;i<n;i++)
            //System.out.println(c[i]);
            count+=c[n-1];
            for(i=n-2;i>=0;i--)
            {
                if(c[i]<=0)
                break;
                if(c[i]>=c[i+1])
                {
                    count+=c[i+1]-1;
                    c[i]=c[i+1]-1;
                    //System.out.println(c[i]-1);
                }
                else if(c[i]<c[i+1])
                {
                    count+=c[i];
                    prev=c[i];
                    //System.out.println(c[i]);
                }
                if(c[i]<=1)
                break;
            }
            System.out.println(count);
        }
    }
}


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;
    }
}