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.
x, y, m = map(int, input().split())
import sys
cnt = 0
if max(x, y) >= m:
    print(0)
    sys.exit()
if max(x, y) <= 0:
    print(-1)
    sys.exit()
if min(x, y) < 0:
    x,y = min(x, y), max(x, y)
    diff = abs(x)
    cnt += diff//y
    x += (diff//y) * y
 
while max(x,y) < m:
    if x > y:
        x, y = y, x
    x += y
    cnt += 1
print(cnt) if max(x, y) >= m else print(-1)