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def rotateImage(a):
    w = len(a)
    h = w
    img =[0]*h
    for col in range(h):
        img_row = [0]*w
        for row in range(w):
            img_row[h-row-1] = a[row][col]
            
        img[col] = img_row

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def sudoku2(grid):
    # test row
    tab = [[False for col in range(9)] for row in range(9)]
   
    cols = [[False for col in range(9)] for row in range(9)]
    rows = [[False for col in range(9)] for row in range(9)]
    for r in range(9):
        for c in range(9):
            if grid[r][c] == ".":
                continue

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def decrypt(word, dictionary):
    lst =[]
    for c in word:
        lst.append(dictionary[c])
    return ''.join(lst)

def isCryptSolution(crypt, solution):
    dict = {}
    for r in solution:
        dict[r[0]] = r[1]

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# Singly-linked lists are already defined with this interface:
# class ListNode(object):
#   def __init__(self, x):
#     self.value = x
#     self.next = None
#
def addTwoHugeNumbers(a, b):
    lst_a = {}
    lst_b = {}
    i=0

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# Singly-linked lists are already defined with this interface:
# class ListNode(object):
#   def __init__(self, x):
#     self.value = x
#     self.next = None
#
def mergeTwoLinkedLists(l1, l2):
    
    if l1 == None and l2 == None:
        return None

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# Definition for singly-linked list:
# class ListNode(object):
#   def __init__(self, x):
#     self.value = x
#     self.next = None
#
def rearrangeLastN(l, n):
    if n == 0:
        return l
    front, back = l, l

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def reverseList(head, tail):
    prev = None
    while prev != tail:
        prev, prev.next, head = head, prev, head.next
    return prev
    
def reverseNodesInKGroups(l, k):
    if k < 2:
        return l
    

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#
# Binary trees are already defined with this interface:
# class Tree(object):
#   def __init__(self, x):
#     self.value = x
#     self.left = None
#     self.right = None

def isEqual(left, right):
    if left == None and right ==  None:

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#
# Binary trees are already defined with this interface:
# class Tree(object):
#   def __init__(self, x):
#     self.value = x
#     self.left = None
#     self.right = None
def kthSmallestInBST(t, k):
        def inorder(r):
                return inorder(r.left) + [r.value] + inorder(r.right) if r else []

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def findProfession(level, pos):
    # Base case 
    if (level == 1): 
        return 'Engineer'
  
    # Recursively find parent's profession. If parent 
    # is a doctar, this node will be a doctal if it is 
    # at odd position and an engineer if at even position 
    if (findProfession(level-1, (pos+1)//2) == 'Doctor'): 
        if (pos%2):