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.
import math

a, b, c = map (int, input().split())

if (max (a, b) >= c):
    print (0)
    raise SystemExit

if (a <= 0 and b <= 0) :
    print (-1)
    raise SystemExit

tot = 0

if ((a <= 0 and b > 0) or (b <= 0 and a > 0)) :
    add = max (a, b)
    menor = min (a, b)
    adicionar = math.ceil(-menor / add)
    tot = adicionar
    
    if (min (a, b) == a) :
        a += add * adicionar
    else :
        b += add * adicionar

times = 500

while (times > 0) :
    times -= 1
    if (max(a, b) >= c) :
        print (tot)
        raise SystemExit
        
    tot += 1
    add = a + b
    if (min (a, b) == a) :
        a = add
    else :
        b = add

print (-1)
