


'''simple docstring'''







lowerCAmelCase    :     List[Any]					=						[
    (10_00, '''M'''),
    (9_00, '''CM'''),
    (5_00, '''D'''),
    (4_00, '''CD'''),
    (1_00, '''C'''),
    (90, '''XC'''),
    (50, '''L'''),
    (40, '''XL'''),
    (10, '''X'''),
    (9, '''IX'''),
    (5, '''V'''),
    (4, '''IV'''),
    (1, '''I'''),
]






def 			lowercase				(_A					):







       """simple docstring"""







       _lowerCAmelCase    :							Any   			=       {'I': 1, 'V': 5, 'X': 1_0, 'L': 5_0, 'C': 1_0_0, 'D': 5_0_0, 'M': 1_0_0_0}
       _lowerCAmelCase    :							Union[str, Any]   			=       0
       _lowerCAmelCase    :							str   			=       0
       while place < len(UpperCAmelCase__					):
              if (place + 1 < len(UpperCAmelCase__					)) and (vals[roman[place]] < vals[roman[place + 1]]):
                     total += vals[roman[place + 1]] - vals[roman[place]]
                     place += 2
              else:
                     total += vals[roman[place]]
                     place += 1
       return total






def 			lowercase				(_A					):







       """simple docstring"""







       _lowerCAmelCase    :							str   			=       []
       for arabic, roman in ROMAN:
              ((_lowerCAmelCase)			,     (_lowerCAmelCase))    :							int   			=       divmod(UpperCAmelCase__   ,						UpperCAmelCase__					)
              result.append(roman * factor					)
              if number == 0:
                     break
       return "".join(UpperCAmelCase__					)


if __name__ == "__main__":
       import doctest

       doctest.testmod()





'''simple docstring'''







lowerCAmelCase    :     Optional[int]					=						[sum(int(c, 10) ** 2 for c in i.__str__()) for i in range(10_00_00)]






def 			lowercase				(_A					):







       """simple docstring"""







       _lowerCAmelCase    :							str   			=       0
       while number:
              # Increased Speed Slightly by checking every 5 digits together.
              sum_of_digits_squared += DIGITS_SQUARED[number % 1_0_0_0_0_0]
              number //= 1_0_0_0_0_0

       return sum_of_digits_squared


# There are 2 Chains made,
# One ends with 89 with the chain member 58 being the one which when declared first,
# there will be the least number of iterations for all the members to be checked.

# The other one ends with 1 and has only one element 1.

# So 58 and 1 are chosen to be declared at the starting.

# Changed dictionary to an array to quicken the solution
lowerCAmelCase    :     list[bool | None]					=						[None] * 10_00_00_00
lowerCAmelCase    :     List[str]					=						True
lowerCAmelCase    :     Union[str, Any]					=						False






def 			lowercase				(_A					):







       """simple docstring"""







       if CHAINS[number - 1] is not None:
              return CHAINS[number - 1]  # type: ignore

       _lowerCAmelCase    :							Any   			=       chain(next_number(_A					)					)
       _lowerCAmelCase    :							List[str]   			=       number_chain

       while number < 1_0_0_0_0_0_0_0:
              _lowerCAmelCase    :							Tuple   			=       number_chain
              number *= 1_0

       return number_chain






def 			lowercase				(_A = 1_0_0_0_0_0_0_0					):







       """simple docstring"""







       for i in range(1   ,						_A					):
              if CHAINS[i] is None:
                     chain(i + 1					)

       return CHAINS[:number].count(_A					)


if __name__ == "__main__":
       import doctest

       doctest.testmod()
       print(F'''{solution() = }''')

