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:
T = int(input())
for _ in range(T) :
    n = int(input())
    xx = 0
    ma = {}; vis = {}
    v = []
    pre = [int(i) for i in input().split()]
    for i in range(n) :
        if pre[i] not in ma.keys() :
            ma[pre[i]] = 1
            v.append(pre[i])
        else :
            ma[pre[i]] += 1
        xx = max(xx, ma[pre[i]])
    for i in v :
        if i not in ma.keys(): continue
        if ma[i] in vis.keys() :
            vis[ma[i]] += 1
        else :
            vis[ma[i]] = 1
    ans = 0; cnt = 0
    # print(xx)
    for i in range(xx, 0, -1) :
        if i in vis.keys() :
            ans += i
            cnt += vis[i] - 1
        else :
            if cnt > 0 :
                ans += i
                cnt -= 1
    print(ans)