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 PYTHON):

Solution:
q = int(raw_input())
for _ in range(q):
    n = int(raw_input())
    a = list(map(int,raw_input().split()))
    d = {}
    for i in range(1,n+1):
        d[i] = 0
    for i in range(len(a)):
        d[a[i]]+=1
    c = []
    for key in d:
        c.append([d[key],key])
    c = sorted(c)
    c.reverse()
    p = c[0][0]
    l = c[0][0]
    for i in range(1,len(c)):
        if(l==1 or c[i][0]==0):
            break
        if(l>c[i][0]):
            p+=c[i][0]
            l = c[i][0]
        elif(l==c[i][0]):
            p+=(c[i][0]-1)
            l = c[i][0]-1
        elif(l<c[i][0]):
            p+=(l-1)
            l = l-1
    print(p)