





from bisect import bisect
from itertools import accumulate




def _a (  SCREAMING_SNAKE_CASE__      :  Dict			,				SCREAMING_SNAKE_CASE__      :  Any			,				SCREAMING_SNAKE_CASE__      :  Dict			,				SCREAMING_SNAKE_CASE__      :  int       )	->    Dict:

       '''simple docstring'''






       SCREAMING_SNAKE_CASE__       :      Tuple     =						sorted(zip(SCREAMING_SNAKE_CASE__			,				SCREAMING_SNAKE_CASE__       )			,				key=lambda SCREAMING_SNAKE_CASE__       : x[0] / x[1]			,				reverse=SCREAMING_SNAKE_CASE__       )
       SCREAMING_SNAKE_CASE__       :      Optional[int]     =						[i[0] for i in r], [i[1] for i in r]
       SCREAMING_SNAKE_CASE__       :      int     =						list(accumulate(SCREAMING_SNAKE_CASE__       )       )
       SCREAMING_SNAKE_CASE__       :      List[Any]     =						bisect(SCREAMING_SNAKE_CASE__			,				SCREAMING_SNAKE_CASE__       )
       return (
           0
           if k == 0
           else sum(vl[:k]       ) + (w - acc[k - 1]) * (vl[k]) / (wt[k])
           if k != n
           else sum(vl[:k]       )
       )


if __name__ == "__main__":
  import doctest

  doctest.testmod()













_lowerCamelCase :    dict[tuple[int, int, int], int]  							= {}




def _a (  SCREAMING_SNAKE_CASE__      :  int			,				SCREAMING_SNAKE_CASE__      :  int			,				SCREAMING_SNAKE_CASE__      :  int       )	->    int:

							'''simple docstring'''






							if late == 3 or absent == 2:
														return 0

							# if we have no days left, and have not failed any other rules,
							# we have a prize string
							if days == 0:
														return 1

							# No easy solution, so now we need to do the recursive calculation

							# First, check if the combination is already in the cache, and
							# if yes, return the stored value from there since we already
							# know the number of possible prize strings from this point on
							SCREAMING_SNAKE_CASE__       :      Union[str, Any]     =						(days, absent, late)
							if key in cache:
														return cache[key]

							# now we calculate the three possible ways that can unfold from
							# this point on, depending on our attendance today

							# 1) if we are late (but not absent), the "absent" counter stays as
							# it is, but the "late" counter increases by one
							SCREAMING_SNAKE_CASE__       :      Tuple     =						_calculate(days - 1			,				SCREAMING_SNAKE_CASE__			,				late + 1       )

							# 2) if we are absent, the "absent" counter increases by 1, and the
							# "late" counter resets to 0
							SCREAMING_SNAKE_CASE__       :      Dict     =						_calculate(days - 1			,				absent + 1			,				0       )

							# 3) if we are on time, this resets the "late" counter and keeps the
							# absent counter
							SCREAMING_SNAKE_CASE__       :      Any     =						_calculate(days - 1			,				SCREAMING_SNAKE_CASE__			,				0       )

							SCREAMING_SNAKE_CASE__       :      str     =						state_late + state_absent + state_ontime

							SCREAMING_SNAKE_CASE__       :      Optional[int]     =						prizestrings
							return prizestrings




def _a (  SCREAMING_SNAKE_CASE__      :  int = 30       )	->    int:

							'''simple docstring'''






							return _calculate(SCREAMING_SNAKE_CASE__			,				absent=0			,				late=0       )


if __name__ == "__main__":
		print(solution())






