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 merge(arr, l, m, r): 
    n1 = m - l + 1
    n2 = r- m 
  
    # create temp arrays 
    L = [0] * (n1) 
    R = [0] * (n2) 
  
    # Copy data to temp arrays L[] and R[] 
    for i in range(0 , n1): 
        L[i] = arr[l + i] 
  
    for j in range(0 , n2): 
        R[j] = arr[m + 1 + j] 
  
    # Merge the temp arrays back into arr[l..r] 
    i = 0     # Initial index of first subarray 
    j = 0     # Initial index of second subarray 
    k = l     # Initial index of merged subarray 
  
    while i < n1 and j < n2 : 
        if L[i] <= R[j]: 
            arr[k] = L[i] 
            i += 1
        else: 
            arr[k] = R[j] 
            j += 1
        k += 1
  
    # Copy the remaining elements of L[], if there 
    # are any 
    while i < n1: 
        arr[k] = L[i] 
        i += 1
        k += 1
  
    # Copy the remaining elements of R[], if there 
    # are any 
    while j < n2: 
        arr[k] = R[j] 
        j += 1
        k += 1
  
# l is for left index and r is right index of the 
# sub-array of arr to be sorted 
def mergeSort(arr,l,r): 
    if l < r: 
  
        # Same as (l+r)//2, but avoids overflow for 
        # large l and h 
        m = (l+(r-1))//2
  
        # Sort first and second halves 
        mergeSort(arr, l, m) 
        mergeSort(arr, m+1, r) 
        merge(arr, l, m, r) 
t=int(input())
for i in range(t):
    n=int(input())
    arr=list(map(int,input().split()))
    count=[0]*n
    for j in range(n):
        count[arr[j]-1]+=1
    realCount=[]
    for j in range(n):
        if count[j]!=0:
            realCount.append(count[j])
    l=len(realCount)
    mergeSort(realCount,0,l-1)
    answer=[realCount[-1]]
    ans=0
    for j in range(2,l+1):
        if realCount[-j]>=answer[-1]:
            if answer[-1]-1!=0:
                answer.append(answer[-1]-1)
        else:
            answer.append(realCount[-j])
    print(sum(answer))