#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  long long cnt = 0;
  if (x >= m || y >= m) {
    cout << cnt;
    return 0;
  }
  if (m < 0 || (m && x <= 0 && y <= 0) || (!m && x < 0 && y < 0)) {
    cout << -1;
    return 0;
  }
  if (x > y) swap(x, y);
  if (x < 0 && y > 0) {
    long long c = (x * -1) / y;
    x = x + (c * y);
    cnt += c;
  }
  while (1) {
    if (x >= m || y >= m) {
      cout << cnt;
      return 0;
    }
    cnt++;
    if (x > y) swap(x, y);
    x = (x + y);
  }
  cout << cnt;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  long long cnt = 0;
  if (x >= m || y >= m) {
    cout << cnt;
    return 0;
  }
  if (m < 0 || (m && x <= 0 && y <= 0) || (!m && x < 0 && y < 0)) {
    cout << -1;
    return 0;
  }
  if (x > y) swap(x, y);
  if (x < 0 && y > 0) {
    long long c = (x * -1) / y;
    x = x + (c * y);
    cnt += c;
  }
  while (1) {
    if (x >= m || y >= m) {
      cout << cnt;
      return 0;
    }
    cnt++;
    if (x > y) swap(x, y);
    x = (x + y);
  }
  cout << cnt;
}

```