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.
from sys import stdin,stdout
from math import ceil
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
def fn(a,b):
    op=0
    while max(a,b)<m:
        a1=max(a,b)+min(a,b)
        b1=max(a,b)
        a,b=a1,b1
        op+=1
    return op
for _ in range(1):#nmbr()):
    a,b,m=lst()
    if a>b:a,b=b,a
    if max(a,b)>=m:
        print(0)
        continue
    if a<=0 and b<=0:
        if (a+b)>=m:
            print(0)
        else:print(-1)
        continue
    if a>0 and b>0:
        print(fn(a,b))
        continue
    if a<=0 and b>0:
        terms=ceil(-a/b)
        start=a+terms*b
        print(terms+fn(start,b))
        continue
    print(-1)