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, raw_input().strip().split(' '))

if x <= 0 and y <= 0: # If we cannot grow it must be good instantly
    if max(x,y) < m:
        print -1
    else:
        print 0
    exit()

if min(x,y) < 0 and max(x,y) > 0 and max(x,y) < m: # If one is < 0 we just make it at least 0 if max < m
    mi = -min(x,y)
    ma = max(x,y)
    if mi % ma == 0:
        counter = mi / ma
    else:
        counter = mi / ma + 1
    if x < y:
        x += counter * ma
    else:
        y += counter * ma
else: # Normal case aka one >0 and the other >=0
    counter = 0

while max(x,y) < m:
    if x < y:
        x = x + y
    else:
        y = x + y
    counter += 1

print counter
