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 {
	public static void main(String[] args) {
		MyScanner in = new MyScanner();
		
		int q=in.nextInt();
		while(q-->0) {
			int n=in.nextInt();
			int[] cnt=new int[n+1];
			for(int i=0; i<n; i++)
				cnt[in.nextInt()]++;
			Arrays.parallelSort(cnt);
			for(int i=0; i<cnt.length/2; i++) {
				int t=cnt[i];
				cnt[i]=cnt[cnt.length-1-i];
				cnt[cnt.length-1-i]=t;
			}
			int prev=cnt[0];
			int ans=prev;
			for(int i=1; i<=n; i++) {
				if(prev==0)
					break;
				if(cnt[i]>=prev) {
					prev=cnt[i]-(cnt[i]-prev)-1; //distance from cnt[i] (above) to prev, then decrement it
					ans+=prev;
				} else {
					prev=cnt[i];
					ans+=prev;
				}
				//2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1s
				//1111 2222 3333 4444
				//4 3 2 1
			}
			System.out.println(ans);
		}
	}

	static class MyScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		String next() {
			while(st==null||!st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
}