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:
def ii():return int(input())
def mi():return map(int,input().split())
def li():return list(mi())
def f(n):
    s=str(n)
    c=0
    for i in s:
        c+=int(i)
    if(c%4==0):
        return 1
    else:
        return 0
t=ii()
while(t):
    t-=1
    n=ii()
    m={}
    a=li()
    for i in range(n):
        if a[i] not in m:
            m[a[i]]=1
        else:
            m[a[i]]+=1
    s=[]
    for i in m.values():
        s.append(i)
    del(m)
    m={}
    for i in s:
        if i not in m:
            m[i]=1
        else:
            m[i]+=1
    b=[]
    c=[]
    for i in s:
        if(m[i]==1):
            b.append(i)
        else:
            c.append(i)
    c.sort()
    m=1
    for i in c:
        for j in range(i,0,-1):
            if j not in b:
                b.append(j)
                break
    print(sum(b))