
from itertools import count





def      lowerCamelCase_					(							lowerCamelCase__ = 5_0       ):
				lowerCamelCase_         =     [1] * min_block_length

				for n in count(lowerCamelCase__       ):
								fill_count_functions.append(1       )

								for block_length in range(lowerCamelCase__      ,					n + 1       ):
												for block_start in range(n - block_length       ):
																fill_count_functions[n] += fill_count_functions[
																    n - block_start - block_length - 1
																]

												fill_count_functions[n] += 1

								if fill_count_functions[n] > 1_0_0_0_0_0_0:
												break

				return n


if __name__ == "__main__":
			print(F"""{solution() = }""")




import math





def      lowerCamelCase_					(							lowerCamelCase__      ,					lowerCamelCase__       ):
				if 0 not in (x, y):
								# We use the relation x^y = y*log10(x), where 10 is the base.
								return y * math.logaa(lowerCamelCase__       )
				else:
								if x == 0:  # 0 raised to any number is 0
												return 0
								elif y == 0:
												return 1  # any number raised to 0 is 1
				raise AssertionError("This should never happen"       )


if __name__ == "__main__":  # Main function
			# Read two numbers from input and typecast them to int using map function.
			# Here x is the base and y is the power.
			__A							='''Enter the base and the power separated by a comma: '''
			__A,		__A							=map(int, input(prompt).split(''','''))
			__A,		__A							=map(int, input(prompt).split(''','''))

			# We find the log of each number, using the function res(), which takes two
			# arguments.
			__A							=res(xa, ya)
			__A							=res(xa, ya)

			# We check for the largest number
			if resa > resa:
						print('''Largest number is''', xa, '''^''', ya)
			elif resa > resa:
						print('''Largest number is''', xa, '''^''', ya)
			else:
						print('''Both are equal''')


