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;
const int N = 90;
long long f[N];
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  if (x > y) swap(x, y);
  if (m <= y) {
    printf("0\n");
    return 0;
  }
  if (y <= 0) {
    printf("-1\n");
    return 0;
  }
  long long res = 0;
  long long k = (-x) / y + 1;
  k = max(0ll, k);
  x = x + k * y;
  f[1] = 1;
  for (int i = 2; i < N; i++) {
    f[i] = f[i - 1] + f[i - 2];
  }
  res += k;
  for (int i = 1; i < N; i++) {
    if (f[i - 1] * x + f[i] * y >= m) {
      break;
    }
    res++;
  }
  cout << res << "\n";
  return 0;
}
