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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int x, y, m;
  cin >> x >> y >> m;
  long long int a, b;
  if ((x >= m) || (y >= m))
    cout << "0\n";
  else if ((x <= 0) && (y <= 0) && (m > 0))
    cout << "-1\n";
  else if ((x < 0) && (y < 0))
    cout << "-1\n";
  else if (m > 0) {
    long long int i = 0;
    if ((x < 0) || (y < 0)) {
      a = min(x, y);
      b = max(x, y);
      a *= -1;
      i = (a / b);
      i++;
      x = b;
      y = (b * i) - a;
    }
    {
      while (1) {
        if ((y >= m) || (x >= m)) {
          cout << i << "\n";
          break;
        }
        i++;
        a = x;
        b = y;
        x = max(a, b);
        y = (a + b);
      }
    }
  }
  return 0;
}
