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 collections import defaultdict
from heapq import *
from sys import stdin, stdout

q = int(stdin.readline())

for it in range(q):
    n = int(stdin.readline())
    d = [0]*n
    f = [0]*n
    for i in range(n):
        t, b = map(int, stdin.readline().split())
        d[t-1]+=1
        if b == 1:
            f[t-1] += 1

    d = [(x, i) for i, x in enumerate(d)]
    d.sort(reverse=True)
    h = []
#    print(l)
    ans = 0
    ans_f = 0
    cur = n
    idx = 0
    while cur > 0:
        while idx < len(d) and d[idx][0] == cur:
            d1 = d[idx]
            heappush(h, (-f[d1[1]], d1[1]))
            idx += 1
        if h:
            ans += cur
            e = heappop(h)
            ans_f += min(cur, -e[0])
        cur -= 1

    stdout.write(str(ans)+" "+str(ans_f)+"\n")