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 yo in range(t):
    n=int(input())
    l=input()
    l=l.split(" ")

    d1={}
    d2={}
    #f=[]
    for i in l:
        k=int(i)
        p=d1.get(k,-1)

        if(p==-1):
            d1[k]=1
            #d2[p]=1
            
        else:
            d1[k]+=1
            #d2[p]+=1
    ans=0
    for u,v in d1.items():
        if(d2.get(v,-1)==-1):
            d2[v]=1
        else:
            d2[v]+=1
    d3={}
    for u,v in d2.items():
        k=0
        i=0
        while(i<v):
            if(k>=u):
                break
            if(d3.get(u-k,-1)==-1):
                d3[u-k]=1
                ans+=u-k
                i+=1
            k+=1
    print(ans)