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 inp(dtype=str, strip=True):
    s = input()
    res = [dtype(p) for p in s.split()]
    res = res[0] if len(res) == 1 and strip else res
    return res


def problem1():
    a = inp(int)
    n = -1
    while n < 0:
        s = 0
        tmp = a
        while tmp:
            s += tmp % 10
            tmp = tmp // 10

        if s % 4 == 0:
            n = a

        a += 1

    print(n)


def problem2():
    q = inp(int)
    for _ in range(q):
        n, k = inp(int)
        a = inp(int, strip=False)

        mn, mx = 0, 10 ** 10
        for price in a:
            mx = min(price + k, mx)
            mn = max(price - k, mn)

        if mn <= mx:
            print(mx)
        else:
            print(-1)


def problem3():
    q = inp(int)
    for _ in range(q):
        k, n, a, b = inp(int)
        if b * n >= k:
            print(-1)
        elif a * n < k:
            print(n)
        else:
            print((k - b * n - 1) // (a - b))


from collections import Counter


def problem4():
    q = inp(int)
    for _ in range(q):
        n = inp(int)
        a = inp(int, strip=False)
        cnt = Counter(a)
        cnt = sorted(cnt.values(), reverse=True)
        res = 0
        prev = 10 ** 6
        # print(cnt)
        for c in cnt:
            # print(prev, c)
            if c >= prev:
                res += prev - 1
                prev = prev - 1
            else:  # c < cur
                res += c
                prev = c

            if prev <= 1:
                break
        print(res)


def problem5():
    pass


if __name__ == '__main__':
    # problem1()
    # problem2()
    # problem3()
    problem4()
    # problem5()