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
# Your code here!
from collections import defaultdict
import sys
import heapq


def input(): return sys.stdin.readline()


def solution():
    n = int(input())
    candi = [[0, 0] for i in range(n+1)]
    que = []
    for i in range(n):
        taste, youwant = map(int, input().split())
        if youwant == 1:
            candi[taste][0] += 1
        else:
            candi[taste][1] += 1

    for i in range(n+1):
        s = sum(candi[i])
        if s > 0:
            que.append((-s, -candi[i][0], -candi[i][1]))
    heapq.heapify(que)
    ans = 0
    badcnt = 0
    border = -que[0][0]
    stocked = []
    heapq.heapify(stocked)  # where not still used
    ording = 0
    while (que or stocked) and border > 0:
        ording += 1
        while que and que[0][0] <= -border:
            poped_cnt, poped_good_candi, poped_bad_candi = heapq.heappop(que)
            heapq.heappush(stocked, (poped_good_candi, poped_bad_candi))

        if not stocked:
            if que:
                border = -que[0][0]
            continue

        # print("phase: ", ording)
        # print("que: ", que)
        # print("stocked: ", stocked)
        # print("border: ", border)
        # print("\n")
        if stocked:
            good_candi, bad_candi = heapq.heappop(stocked)
            good_candi *= -1
            bad_candi *= -1
            sums = good_candi + bad_candi
            if sums < border:
                ans += sums
                border = sums - 1
                badcnt += good_candi
                continue

            elif sums >= border:
                ans += border
                badcnt += min(border, good_candi)
                border -= 1
                continue

    print(ans, badcnt)

    return


def main():
    test = int(input())
    for i in range(test):
        solution()
    return


if __name__ == "__main__":
    main()