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:
#import collections
# import random
# import math
#import itertools
#import math
import math
from collections import defaultdict
# import itertools
# from sys import stdin, stdout
#import math
import sys
# import operator
# from decimal import Decimal

# sys.setrecursionlimit(10**6)

p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()


def solve():
    n = ii()
    a = li()
    d = defaultdict(int)
    for i in range(n):
        d[a[i]]+=1
    all = []
    #print(d)
    for key in d:
        all.append(d[key])
    all.sort(reverse=True)
    m = len(all)
    #print(all)
    j = 1
    p = all[0]-1
    while j<m:
        if all[j]>=p:
            all[j] = p
            p = p-1 if p>0 else 0
            j+=1
        else:
            p = all[j]-1 if all[j]>0 else 0
            j+=1

    return sum(all)

def main():
    for _ in range(ii()):
        sys.stdout.write(str(solve()) + "\n")


    #     z += str(ans) + '\n'
    # print(len(ans), ' '.join(map(str, ans)), sep='\n')
    # stdout.write(z)


# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()


if __name__ == "__main__":
    main()