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:
from __future__ import division, print_function
import bisect
import math
import itertools
import sys
from atexit import register
from collections import Counter

if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream


if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip


def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.

    Args:
        sync (bool, optional): The new synchronization setting.

    """
    global input, flush

    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')

        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
def main():
    q=int(input())
    for i in range(q):
        n=int(input())
        a=list(map(int,input().split()))
        d=dict(Counter(a))
        ld=len(d)
        psa=[0]*ld
        j=0
        for i in d.values():
            psa[j]=i
            j+=1
        psa=sorted(psa)
        psa=psa[::-1]
        dd=0
        pre=-1
        c=dict()
        for i in range(ld):
            while psa[i] in c and psa[i]>=0:
                psa[i]-=1
            if psa[i]>=0:
                c[psa[i]]=1
                dd+=psa[i]
        print(dd)
    
if __name__ == '__main__':
    sync_with_stdio(False)
    main()