

'''simple docstring'''







from collections import defaultdict
from math import ceil, sqrt
def    UpperCAmelCase__     (   UpperCAmelCase__ = 1_00_00_00,							UpperCAmelCase__ = 10					)      ->      int:
       A_						       =   defaultdict(UpperCAmelCase__					)

       for outer_width in range(3,							(t_limit // 4) + 2					):
              if outer_width * outer_width > t_limit:
                     A_						       =   max(
                         ceil(sqrt(outer_width * outer_width - t_limit					)					),							1					)
              else:
                     A_						       =   1

              hole_width_lower_bound += (outer_width - hole_width_lower_bound) % 2

              for hole_width in range(UpperCAmelCase__,							outer_width - 1,							2					):
                     count[outer_width * outer_width - hole_width * hole_width] += 1

       return sum(1 for n in count.values() if 1 <= n <= 10					)


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



'''simple docstring'''







from math import pow
def    UpperCAmelCase__     (   UpperCAmelCase__,							UpperCAmelCase__,							UpperCAmelCase__,							UpperCAmelCase__,							UpperCAmelCase__,							)      ->      tuple[int, int]:
       if current_sum == needed_sum:
              # If the sum of the powers is equal to needed_sum, then we have a solution.
              solutions_count += 1
              return current_sum, solutions_count

       A_						       =   int(pow(UpperCAmelCase__,							UpperCAmelCase__					)					)
       if current_sum + i_to_n <= needed_sum:
              # If the sum of the powers is less than needed_sum, then continue adding powers.
              current_sum += i_to_n
              A_       ,      A_						       =   backtrack(
                  UpperCAmelCase__,							UpperCAmelCase__,							current_number + 1,							UpperCAmelCase__,							UpperCAmelCase__					)
              current_sum -= i_to_n
       if i_to_n < needed_sum:
              # If the power of i is less than needed_sum, then try with the next power.
              A_       ,      A_						       =   backtrack(
                  UpperCAmelCase__,							UpperCAmelCase__,							current_number + 1,							UpperCAmelCase__,							UpperCAmelCase__					)
       return current_sum, solutions_count
def    UpperCAmelCase__     (   UpperCAmelCase__,							UpperCAmelCase__					)      ->      int:
       if not (1 <= needed_sum <= 10_00 and 2 <= power <= 10):
              raise ValueError(
                  """Invalid input\n"""
                  """needed_sum must be between 1 and 1000, power between 2 and 10."""					)

       return backtrack(UpperCAmelCase__,							UpperCAmelCase__,							1,							0,							0					)[1]  # Return the solutions_count


if __name__ == "__main__":
      import doctest

      doctest.testmod()
