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 PYTHON3):

Solution:
import sys

q=int(sys.stdin.readline())

ans_arr=[]

for x in range(q):
	n=int(sys.stdin.readline())

	a=[int(i) for i in sys.stdin.readline().split()]

	# arr=[0]*(2*pow(10,5)+1)

	# for x in range(n):
	# 	arr[a[x]]+=1

	a.sort()

	arr=[]

	prev=-2
	ind=-1
	for x in range(n):
		if(a[x]!=prev):
			arr.append(1)
			ind+=1
		else:
			arr[ind]+=1

		prev=a[x]
	# d={}

	# for x in range(n):
	# 	if(a[x] not in d):
	# 		d[a[x]]=1
	# 	else:
	# 		d[a[x]]+=1	

	# arr=[]
	# for k in d:
	# 	arr.append(d[k])

	arr.sort()
	answer=0
	last=arr[-1]+1

	for g in range(len(arr)-1,-1,-1):
		val=max(0,min(last-1,arr[g]))
		answer+=val
		last=val

	ans_arr.append(str(answer))

print("\n".join(ans_arr))