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 INF = (int)2e+9;
const int MAX_N = 2e5 + 7;
long long maxl(long long a, long long b) {
  if (a > b) return a;
  return b;
}
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  if (a >= c || b >= c) {
    cout << 0;
    return 0;
  }
  if (a <= 0 && b <= 0) {
    cout << -1;
    return 0;
  }
  long long ans = 0;
  while (a < c && b < c) {
    if (a < 0 || b < 0) {
      if (a < 0) {
        ans += maxl(1, -a / b);
        a += maxl(-a / b, 1) * b;
      } else if (b < 0) {
        ans += maxl(-b / a, 1);
        b += maxl(-b / a, 1) * a;
      }
      continue;
    }
    (a < b) ? a += b : b += a;
    ans++;
  }
  cout << ans;
  return 0;
}
