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.
def solve():
    x, y, m = (input().split())
    x = int(x)
    y = int(y)
    m = int(m)

    if x > y:
        x, y = y, x

    if y >= m:
        print(0)
        return
    if y <= 0:
        print(-1)
        return
    a = 1
    b = 1
    c = 0
    temp = y
    cnt = 0
    while temp < m:
        a = b+c
        c = b
        b = a
        cnt += 1
        temp = a*y + b*x
    print(cnt)


solve()
