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 collections
T=int(input())
for i in range(0,T):
    x=int(input())
    arr=list(map(int,input().split()))
    arr.sort()
    # print(arr)
    arr1=[]
    x=arr[0]
    count=0
    for i in range(1,len(arr)):
        if x==arr[i]:
            count+=1
            continue
        
        p=(x,count+1)
        arr1.append(p)
        x=arr[i]
        count=0
    p=(x,count+1)
    arr1.append(p)
    # print(arr3)
    arr1=sorted(arr1, key=lambda x: x[1])
    # counter=(collections.Counter(arr))
    # arr1=counter.most_common(len(arr))
    # print(arr,arr3)
    # print(arr1)
    x1=arr1[len(arr1)-1][1]
    c=arr1[len(arr1)-1][1]
    for j in range(len(arr1)-2,-1,-1):
        # if c==arr1[j][1] or c<arr1[j][1]:
        #     x1=x1+c-1
        #     c=c-1
        #     if c==1:
        #         break
        # elif c!=arr1[j][1]:
        #     x1=x1+arr1[j][1]
        #     c=arr1[j][1]
        c-=1
        if c==0:
            break
        if c>=arr1[j][1]:
            x1+=arr1[j][1]
            c=arr1[j][1]
        else:
            x1+=c
    arr=[]
    arr1={}
    arr2=[]
    print(x1)