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:
from sys import stdin
input = stdin.readline
q = int(input())
for query in range(q):
	n = int(input())
	l = list(map(int,input().split()))
	l.sort()
	roz = 1
	for i in range(1, n):
		if l[i - 1] != l[i]:
			roz += 1
	d = l + [-1]
	ziom = [0] * roz
	count = 1
	j = 0
	for i in range(1, n + 1):
		if d[i] == d[i - 1]:
			count += 1
		else:
			ziom[j] = count
			j += 1
			count = 1
	ziom.sort()
	thres = 1000000000000000000000
	wyn = 0
	i = len(ziom) - 1
	while True:
		if thres <= 0 or i < 0:
			print(wyn)
			break
		wyn += min(thres - 1, ziom[i])
		thres = min(thres - 1, ziom[i])
		i -= 1