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
input = sys.stdin.readline
import heapq
from collections import Counter

Q=int(input())
for testcase in range(Q):
    n=int(input())
    C=[list(map(int,input().split())) for i in range(n)]

    D=[[0,0] for i in range(n+1)]

    for x,y in C:
        D[x][0]+=1
        if y==1:
            D[x][1]+=1

    W=[]
    for i in range(n+1):
        if D[i][0]!=0:
            W.append((D[i][0],D[i][1]))

    W.sort(reverse=True)
    W.append((0,0))
    #print(W)

    C2=Counter([c[0] for c in C])
    S=sorted(C2.values(),reverse=True)

    NOW=10**10

    ANS=[]
    for s in S:
        M=min(NOW,s)
        ANS.append(M)
        NOW=M-1

        if NOW==0:
            break

    #print(ANS,sum(ANS))

    H=[]
    i=0
    ANS1=0
    for ans in ANS:
        while W[i][0]>=ans:
            heapq.heappush(H,-W[i][1])
            i+=1

        x=min(ans,-heapq.heappop(H))
        ANS1+=x
    print(sum(ANS),ANS1)