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;
inline long long calc(long long xi, long long yi) {
  long long x, y, ans;
  x = min(xi, yi);
  y = max(xi, yi);
  ans = 0ll;
  if (y >= m) return ans;
  ans = calc(y, x + y) + 1;
  return ans;
}
int main() {
  long long x, y, xi, yi, k;
  long long ops;
  ops = 0;
  cin >> x >> y >> m;
  if (x >= m || y >= m)
    cout << ops << endl;
  else {
    if (x <= 0 && y <= 0)
      printf("-1\n");
    else {
      if (x >= 0 && y >= 0) {
        ops = calc(x, y);
      } else {
        xi = min(x, y);
        yi = max(x, y);
        k = (yi - xi) / yi - 1;
        xi = xi + k * yi;
        while (1) {
          if (xi >= 0) break;
          xi += yi;
          k++;
        }
        ops = k + calc(xi, yi);
      }
      cout << ops << endl;
    }
  }
  return 0;
}
