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?

Input

Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).

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.

Output

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

Examples

Input

1 2 5


Output

2


Input

-1 4 15


Output

4


Input

0 -1 5


Output

-1

Note

In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).

In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).

Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.
x, y, m = map(int,input().split())
num = 0
if m > max(x,y):
    if x + y > 0:
        while m > x + y:
            if x > y:
                y = x + y
            else:
                x = x + y
            num += 1
        print(num+1)
    elif x + y > min(x,y):
        if x > y:
                num += abs(y // x) + 1
                y = y % x
        else:
                num += abs(x // y) + 1
                x = x % y
        while m > x + y:
            if x > y:
                y = x + y
            else:
                x = x + y
            num += 1
        print(num)
    else:
        print(-1)
else:
    print(0)
