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;
long long m, x, y;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> x >> y >> m;
  if (x < m && y < m && x <= 0 && y <= 0)
    puts("-1");
  else {
    if (x >= m || y >= m) return 0 * puts("0");
    long long ans = 0;
    if (x > y) swap(x, y);
    if (x < 0) {
      long long tmp = (-x) / y + ((-x) % y == 0 ? 1 : 0);
      ans += tmp;
      x = x + y * tmp;
    }
    while (x < m && y < m) {
      long long tmp = x + y;
      if (x < y)
        x = tmp;
      else
        y = tmp;
      ans++;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
