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 arr[100000];
int main() {
  bool flag = 0;
  long long a, b, c;
  cin >> a >> b >> c;
  long long result = 0;
  if (a >= c || b >= c)
    cout << 0 << endl;
  else if (a + b >= c)
    cout << 1 << endl;
  else if ((a < 0 && b < 0) || (a < 0 && b == 0) || (b < 0 && a == 0) ||
           (a == 0 && b == 0))
    cout << -1 << endl;
  else {
    if (min(a, b) < 0) {
      if (abs(min(a, b)) > max(a, b))
        result += abs(min(a, b)) / max(a, b);
      else
        result += max(a, b) / abs(min(a, b));
      if (min(a, b) == a) {
        a += result * b;
      } else {
        b += result * a;
      }
    }
    if (max(a, b) >= c) {
      cout << result << endl;
      flag = 1;
    }
    arr[0] = max(a, b);
    arr[1] = a + b;
    if (!flag && arr[1] >= c) {
      cout << result + 1 << endl;
      flag = 1;
    }
    result++;
    int x = 1;
    if (!flag) {
      while (1) {
        x++;
        arr[x] = arr[x - 1] + arr[x - 2];
        result++;
        if (arr[x] >= c) break;
        if (arr[x] <= arr[x - 1]) {
          result = -1;
          break;
        }
      }
      cout << result << endl;
    }
  }
  return 0;
}
