Let us call a pair of integer numbers *m*-perfect, if at least one number in the pair is greater than or equal to *m*. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers *x*, *y* are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (*x* + *y*).

What is the minimum number of such operations one has to perform in order to make the given pair of integers *m*-perfect?
Single line of the input contains three integers *x*, *y* and *m* ( - 10<sup class="upper-index">18</sup> ≤ *x*, *y*, *m* ≤ 10<sup class="upper-index">18</sup>).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
# -*- coding: utf-8 -*-

def main():
	x, y, m = map(int, input().split())
	b1 = min(x, y)
	b2 = max(x, y)
	count = 0
	if b2 > m:
		return 0
	if b2 <= 0:
		return -1
	if b1 < 0:
		count = (b2 - b1 + (b2 - 1)) // b2
		b = b2
		b2 = b1 + count * b2
		b1 = b
	while b2 < m:
		b = b1 + b2
		b1 = min(b2, b)
		b2 = max(b2, b)
		count += 1
	return count

if __name__ == '__main__':
	print('{:#d}'.format(main()))