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 x, y, m, a, b, ans = 0, t;
  cin >> x >> y >> m;
  a = max(x, y);
  b = min(x, y);
  if (a >= m)
    cout << "0" << '\n';
  else {
    if (a + b <= b) {
      cout << "-1" << '\n';
    } else {
      ans += (a - b) / a + ((a - b) % a == 0 ? 0 : 1);
      b += ((a - b) / a + ((a - b) % a == 0 ? 0 : 1)) * a;
      if (b > a) {
        t = a;
        a = b;
        b = t;
      }
      while (a < m) {
        ans++;
        b += a;
        if (b > a) {
          t = a;
          a = b;
          b = t;
        }
      }
      cout << ans << '\n';
    }
  }
}
