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.
a,b,m=[int(i) for i in input().split()]
if a > b:a,b = b,a
if a + b <= 0:print(-1);exit(0)
if b >= m:print(0);exit(0)
f1 = 1
f2 = 1
i = 1
while 1:
 if a * f1 + b * f2 >= m:print(i);exit(0)
 i += 1
 f2 += f1
 f1 = f2 - f1