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.
a, b, m = map(int, raw_input().split())
steps = 0

if a >= m or b >= m:
  print 0
else:
  # make positive
  if a <= 0 and b <= 0:
    print -1
  else:
    if a > b:
      a, b = b, a
    if a < 0:
      steps += ((-a / b)-1)
      if steps < 0:
        steps = 0
      a += steps * b
    addstep = 0
    while True:
      if addstep >= 1000:
        print -1
        break
      if a >= m or b >= m:
        print steps
        break
      steps += 1
      addstep += 1
      if a < b:
        a = (a+b)
      else:
        b = (a+b)

