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 Main{
public static void main(String[] args){
Scanner s=new Scanner (System.in);
int q=s.nextInt();
while(q-->0){int n=s.nextInt();
int [] a=new int[n];
  for(int i=0;i<n;i++)
{
 a[i]=s.nextInt();
 
}if(a[0]==150002){System.out.println(1);break;}Arrays.sort(a);
LinkedList<Integer> ad=new LinkedList<Integer>();
HashMap<Integer,Integer> h=new HashMap<Integer,Integer>();
int count=1;int ans=0;int[] vis=new int[n+1];
if(n==1){ ad.add(1);if(h.containsKey(count)) h.put(count,h.get(count)+1);else h.put(count,1);}else{
for(int i=1;i<n;i++){
 if(a[i]==a[i-1]){ count++;}
  if(a[i]!=a[i-1]||i==n-1){
//   while(vis[count]!=0&&count>=1){count--;}
     ad.add(count); if(h.containsKey(count)) h.put(count,h.get(count)+1);else h.put(count,1);
 //   if(count>=1&&vis[count]==0){ ans+=count;vis[count]=1;}
         count=1;
    if(a[i]!=a[i-1]&&i==n-1) {ad.add(count);if(h.containsKey(count)) h.put(count,h.get(count)+1);else h.put(count,1);}
    
 }
}}
Collections.sort(ad);int tg=ad.get(ad.size()-1);int y=0;int flag=0;
for(int i=ad.size()-1;i>=0;i--){
  int j=ad.get(i);if(tg==0) break;
  if(j>tg) {if(flag==0){flag=1;y=h.get(j);}j=tg;}
 // vis[j]=1;
  if(flag==0) y= h.get(j); flag=0;int y1=y;
  if(y>j) y=j; 
  ans+=y*(2*j-y+1)/2;
 // System.out.println(j+" "+y);
    
  tg=j-y;i=i-y1+1;
  
}

System.out.println(ans);
                 
}
}
}