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.
def haha():
    t = raw_input()
    t = t.split(" ")
    x = int(t[0])
    y = int(t[1])
    m = int(t[2])
    t = 0
    # make x >= y
    if y > x:
        t = x 
        x = y 
        y = t
        t = 0
    if x >= m:
        t = 0
    # situations dont need while
    if x <= 0 and x < m:
        t = -1
   
    while y <= 0 and 0 < x < m:
        if (-y)%(x) == 0:
            t = (-y)/(x) + 1
            y = x
        else:
            t = (-y)//x + 1
            y = y + t*x
    while y > 0 and x > 0 and x < m and y < m:
        if x >= y:
            y = x + y
        else: 
            x = x + y
        t+=1
    print t
haha()