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# -.- coding: latin-1 -.-
from math import sqrt
"""
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.
"""
def sum_of_divisors(n):
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):
"""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())))