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 PYTHON):

Solution:
import sys

def main():
    q = int(sys.stdin.readline())


    for each in xrange(q):
        n = int(sys.stdin.readline())

        line = sys.stdin.readline()
        typeCount = [0] * n
        count = [0] * n

        for i in line.split(' '):
            typeCount[int(i)-1] += 1

        for i in xrange(0, n):
            if typeCount[i] != 0:
                count[typeCount[i]-1] += 1

        possibleNumbers = 0
        empty = []

        for i in xrange(0, n):
            if count[i] == 0: empty.append(i)
            elif count[i] == 1: possibleNumbers += i + 1
            else:

                lenempty = len(empty)
                end = lenempty - min(lenempty, count[i] - 1) - 1

                possibleNumbers += i + 1
                for i in xrange(len(empty)-1, end, -1):
                    possibleNumbers += empty.pop() + 1


        print possibleNumbers


main()