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() {
  int t = 1;
  while (t--) {
    long long x, y, m;
    cin >> x >> y >> m;
    long long a, b;
    a = max(x, y);
    b = min(x, y);
    if ((a <= 0 && b <= 0 && m <= a) || (a > 0 && b <= 0 && m <= a) ||
        (a > 0 && b > 0 && m <= a)) {
      cout << "0\n";
    } else if ((a <= 0 && b <= 0 && m > a) || (a > 0 && b > 0 && m <= 0)) {
      cout << "-1\n";
    } else {
      long long ans = 0;
      if ((a > 0 && b < 0 && m > a)) {
        ans = a - b;
        ans = ans / a;
        ans--;
      } else if ((a > 0 && b >= 0 && m > a)) {
        ans = 0;
      }
      if (b < 0) {
        b = ans * a + b;
      }
      long long count = 0;
      while (a < m && b < m) {
        if (a < b) {
          a = a + b;
          count++;
        } else {
          b = b + a;
          count++;
        }
      }
      cout << count + ans << endl;
    }
  }
  return 0;
}
