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:
# @author 

import sys

class DCandyBoxEasyVersion:
    def solve(self):
        q = int(input())
        for _ in range(q):
            n = int(input())
            a = [int(_) for _ in input().split()]
            d = {key: 0 for key in a}
            for x in a:
                d[x] += 1
            rev_d = {key: 0 for key in d.values()}
            for x in d:
                rev_d[d[x]] += 1

            v = [x for x in rev_d]
            v.sort()

            cur = max(rev_d)
            cnt = max(rev_d)
            ans = 0
            while 1:
                if cnt == 0 or cur == 0:
                    break
                if cur > cnt:
                    cur -= 1
                    continue
                if cnt not in rev_d or rev_d[cnt] == 0:
                    cnt -= 1
                    continue
                rev_d[cnt] -= 1
                ans += cur
                cur -= 1
                # rev_d[cnt - cur] += 1
            print(ans)




solver = DCandyBoxEasyVersion()
input = sys.stdin.readline

solver.solve()