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.util.ArrayList;
public class cf_3
{
	public static void main(String[]args)
	{
		Scanner sc = new Scanner(System.in);
		int q = sc.nextInt();
		for(int i=0;i<q;i++)
		{
			int n = sc.nextInt();
			//HasMap <Integer,Integer> map = new HasMap<Integer,Integer>();
			int arr[] = new int[n];
			for(int j=0;j<n;j++){
				arr[sc.nextInt()-1] += 1; 
			}
			Arrays.sort(arr);
			// for(int i1=0;i1<n;i1++){
			// 	System.out.print(arr[i1]+" ");
			// }
			// int t =0;
			// for(int u=0;u<n;u++){
			// 	if(arr[u]>0){
			// 		t=u;
			// 		break;
			// 	}
			// }
			int val=arr[n-1];
			int sum=val;
			// for(int i1=0;i1<n-t-1;i1++){
			// 	val=val-1;
			// 	if(val>0){
			// 		sum = sum+val;
			// 	}
			// 	else{
			// 		break;
			// 	}
			// }
			// System.out.println(sum);
			for(int j=n-2;j>=0;j--){
				val =val-1;
				if(val>arr[j]){
					val =arr[j];

				}
				if(val<=0){
					break;
				}
			//	System.out.println(val);
				sum=sum+val;
			}
			System.out.println(sum);
			// ArrayList<Integer> arr1 = new ArrayList<Integer>();
			// arr1.add(arr[n-1]);
			// int x =1;
			// for(int j=n-2;j>=0;j--){
			// 	if(arr1.indexOf(arr[j])!=-1){
			// 		if((arr[j]-x)<1){
			// 			break;
			// 		}
			// 		else if((arr[j]-x)==1){
			// 			arr1.add(arr[j]-x);
			// 			break;
			// 		}
			// 		else if((arr[j]-x)>1){
			// 		  arr1.add(arr[j]-x);
			//         }
			//         x++;
					
			// 	}
			// 	else if(arr1.indexOf(arr[j])==-1){
			// 		x=1;
			// 		if(arr[j]>1){
			// 			arr1.add(arr[j]);
			// 		}
			// 		else if(arr[j]==1){
			// 			arr1.add(arr[j]);
			// 			break;
			// 		}
			// 		else if(arr[j]<1){
			// 			break;
			// 		}
			// 	}
			    
			// }
			// for(int o=0;o<arr1.size();o++){
			// 	//System.out.print(arr1.get(o)+" ");
			// 	sum=sum+arr1.get(o);
			// }
			// System.out.println(sum);
		}
	}
}