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:
# -*- coding: utf-8 -*-
"""
Created on Mon Jul  1 09:38:35 2019

@author: dake3
"""
from collections import Counter

q = int(input())
prints = []
for i in range(q):
    n = int(input())
    a = list(map(int, input().split()))
    '''
    b = {}
    for i in range(length):
        b[i] = a.count(i)
    for i in range(length):
        if b[i] == 0:
            del b[i]
    '''
    b = Counter(a)
    listA = list(b.values())
    listA.sort()
    listA.reverse()
    total = 0
    
    for i in range(1, len(listA)):
        if listA[i] >= listA[i-1]:
            listA[i] = listA[i-1]-1

    x = 0
    for i in listA:
        if i>0:
            x+=i
    
    prints.append(str(x))
    #print(x)
print("\n".join(prints))
 
'''
    def runA(listA):
        listC = []
        for listA in listA:
            if listA not in listC:
                listC.append(listA)
            else:
                listC.append(listA-1)
        return listC
    
    def Test(listA):
        if len(listA) != len(set(listA)):
            return False
        else:
            return True
        
    while Test(listA) == False:
        listA = runA(listA)
    
    x = 0
    for i in listA:
        if i>0:
            x+=i
'''