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
from heapq import heappush, heappop

if __name__ == '__main__':
    q = int(input())
    lines = stdin.readlines()
    ans = []
    j = 0
    for _ in range(q):
        n = int(lines[j])
        cnt = {}
        j += 1
        for t in range(n):
            a, f = map(int, lines[j].split())
            if a not in cnt:
                cnt[a] = [1, f]
            else:
                cnt[a][0] += 1
                cnt[a][1] += f

            j += 1
        vls = list(cnt.values())
        vls.sort(reverse=True)
        hp = []
        x = 10**19
        s = 0
        fs = 0
        i = 0
        while x > 0 and (i < len(vls) or hp):
            x = x - 1 if hp else min(x - 1, vls[i][0])

            while i < len(vls) and x <= vls[i][0]:
                heappush(hp, ((-1)*vls[i][1], vls[i][0]))
                i += 1

            f, _ = heappop(hp)
            s += x
            fs += min((-1)*f, x)

        ans.append( "%s %s" % (s, fs))

    print('\n'.join(ans))