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"""

Amicable Numbers

Problem 21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n

which divide evenly into n).

If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and

each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55

and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and

142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

"""

from math import sqrt

def sum_of_divisors(n: int) -> int:

total = 0

for i in range(1, int(sqrt(n) + 1)):

if n % i == 0 and i != sqrt(n):

total += i + n // i

elif i == sqrt(n):

total += i

return total - n

def solution(n: int = 10000) -> int:

"""Returns the sum of all the amicable numbers under n.

>>> solution(10000)

31626

>>> solution(5000)

8442

>>> solution(1000)

504

>>> solution(100)

0

>>> solution(50)

0

"""

total = sum(

i

for i in range(1, n)

if sum_of_divisors(sum_of_divisors(i)) == i and sum_of_divisors(i) != i

)

return total

if __name__ == "__main__":

print(solution(int(str(input()).strip())))