#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  if (x < y) swap(x, y);
  vector<long long> fibo = {1, 1};
  while (1) {
    long long add = fibo[(int)fibo.size() - 1] + fibo[(int)fibo.size() - 2];
    if (add < 0) break;
    fibo.push_back(add);
  }
  long long cnt;
  if (x >= m || y >= m) {
    cnt = 0;
  } else {
    if (x <= 0 && y <= 0) {
      cnt = -1;
    } else {
      cnt = 0;
      if (x > 0 && y <= 0) {
        if (y < 0) {
          y *= -1;
          long long div = (y + x - 1) / x;
          y = div * x - y;
          cnt = div;
        }
        if (y == 0) {
          y += x;
          cnt++;
        }
      }
      for (int i = 0; i < (int)fibo.size() - 1; i++) {
        if (fibo[i] * y + fibo[i + 1] * x >= m) {
          cnt += (i + 1);
          break;
        }
      }
    }
  }
  cout << cnt << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
  if (x < y) swap(x, y);
  vector<long long> fibo = {1, 1};
  while (1) {
    long long add = fibo[(int)fibo.size() - 1] + fibo[(int)fibo.size() - 2];
    if (add < 0) break;
    fibo.push_back(add);
  }
  long long cnt;
  if (x >= m || y >= m) {
    cnt = 0;
  } else {
    if (x <= 0 && y <= 0) {
      cnt = -1;
    } else {
      cnt = 0;
      if (x > 0 && y <= 0) {
        if (y < 0) {
          y *= -1;
          long long div = (y + x - 1) / x;
          y = div * x - y;
          cnt = div;
        }
        if (y == 0) {
          y += x;
          cnt++;
        }
      }
      for (int i = 0; i < (int)fibo.size() - 1; i++) {
        if (fibo[i] * y + fibo[i + 1] * x >= m) {
          cnt += (i + 1);
          break;
        }
      }
    }
  }
  cout << cnt << endl;
  return 0;
}

```