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().strip().split())
x,y = min(x,y),max(x,y)
if y>= m:
  print(0)
elif y<=0:
  print(-1)
elif x>=0:
  i = 0
  while y<m:
    x,y = min(x+y,y) , max(x+y,y)
    i += 1
  print(i)
else:
  n = (y-x)//y
  if n*y + x < y:
    n += 1
  if n*y + x >= m:
    print(n)
  else:
    i = n
    x,y = y , n*y + x
    while y<m:
      x,y = min(x+y,y) , max(x+y,y)
      i += 1
    print(i)
