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:
def addInDict(x, y):
    global d, d2
    if x not in d:
        d[x] = 0
    if x not in d2[y]:
        d2[y][x] = 0
    d[x] += 1
    d2[y][x] += 1

def calcSum1():
    global d, d2, d3, d4, d4Keys, s1, s2
    for kol in d4Keys:
        for numPart in d4[kol]:
            kol1 = d[numPart]
            while kol1 in d3 and kol1 > 0:
                kol1 -= 1
            if kol1 > 0:
                d3[kol1] = 1
                s1 += kol1
                s2 += min(kol, kol1)
                if numPart in d2[0]:
                    del d2[0][numPart]
            
            
def calcSum0():
    global d, d2, d3, s1, s2
    for key in d2[0]:
        kol = d[key]
        while kol in d3 and kol > 0:
            kol -= 1
        if kol > 0:
            d3[kol] = 1
            s1 += kol

import sys
for _ in range(int(input())):
    n = int(sys.stdin.readline())
    d, d2, d3, d4 = {}, {}, {}, {}
    d2[0] = {}
    d2[1] = {}
    for __ in range(n):
        a, f = map(int, sys.stdin.readline().split())
        addInDict(a, f)
##    print('d=', d)
##    print('d2=', d2)
    s1, s2 = 0, 0
    for i in d2[1]:
        if d2[1][i] not in d4:
            d4[d2[1][i]] = {}
        d4[d2[1][i]][i] = i
##    print('d4=', d4)
    d4Keys = list(d4.keys())
    d4Keys.sort(reverse = True)
##    print('d4Keys=', d4Keys)
    
    calcSum1()
    calcSum0()
    print(s1, s2)