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:
for case in range (int (input ())) :
    n = int (input())
    a = list (map (int, input ().split ()))

    ha = {}
    p = 0
    cnt = []
    sta = [0]
    mx = 0

    for x in a :
        if not x in ha.keys() :
            ha[x] = p
            p += 1
            cnt.append(1)
            continue
        cnt[ha[x]] += 1
    
    cnt.sort()

    s = 0

    for x in cnt :
        if x > mx :
            for y in range (mx + 1, x + 1) :
                sta.append(y)

        l, r = 0, len (sta) - 1
        while l <= r :
            mid = (l + r) >> 1
            if sta[mid] <= x :
                l = mid + 1
            else :
                r = mid - 1
        s += sta[r]

        if r :
            sta.pop(r)

        mx = max (x, mx)
            
    print (s)