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())
def proc(x,y,m):
    if max(x, y) >= m:
        print(0)
        exit()
    if max(x, y) <= 0:
        print(-1)
        exit()
    count = 0
    x, y = max(x, y), min(x, y)
    count+=((x-y)//x)
    y+=count*x
    while max(x, y) < m:
        x, y = max(x, y), min(x, y)  # 默认x大y小
        y += x
        count += 1
    print(count)
proc(x,y,m)