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:
import sys
for _ in range(int(input())):
    n=int(sys.stdin.readline())
    a=list(map(int,sys.stdin.readline().split()))
    d={}
    for i in range(n):
        if a[i] in d:
            d[a[i]]+=1
        else:
            d[a[i]]=1
    #print(d)
    b=list(dict.values(d))
    c=[0]*n
    ans=0
    if b[0]==n:
        print(n)
    else:
        for i in range(len(b)-1,-1,-1):
            if c[b[i]]==0:
                ans+=b[i]
                c[b[i]]=1
                #print("adfasf")
            else:
                while c[b[i]]==1:
                    b[i]-=1
                c[b[i]]=1
                c[0]=0
                ans+=b[i]
            #print(b[i])
        print(ans)