#include <bits/stdc++.h>
using namespace std;
long long a, b, c, d, e, f, g, h, i, j, k, m, n;
void run(long long x, long long y, long long m) {
  long long count = 0;
  while (x < m && y < m) {
    if (x <= 0 && y <= 0) {
      count = -1;
      break;
    }
    if (x < 0) {
      if (m < 0)
        g = m;
      else
        g = 0;
      j = (g - x) / y;
      k = (g - x) % y;
      count += j + (k ? 0 : 1);
      x = k ? g : (g - k + y);
    } else if (y < 0) {
      if (m < 0)
        g = m;
      else
        g = 0;
      j = (g - y) / x;
      k = (g - y) % x;
      count += j + (k == 0 ? 0 : 1);
      y = k == 0 ? g : (g - k + x);
    } else if (x > y) {
      y = x + y;
      count++;
    } else if (y >= x) {
      x = x + y;
      count++;
    }
  }
  cout << count << endl;
}
int main() {
  cin >> a >> b >> c;
  run(a, b, c);
  return 0;
}

### 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;
long long a, b, c, d, e, f, g, h, i, j, k, m, n;
void run(long long x, long long y, long long m) {
  long long count = 0;
  while (x < m && y < m) {
    if (x <= 0 && y <= 0) {
      count = -1;
      break;
    }
    if (x < 0) {
      if (m < 0)
        g = m;
      else
        g = 0;
      j = (g - x) / y;
      k = (g - x) % y;
      count += j + (k ? 0 : 1);
      x = k ? g : (g - k + y);
    } else if (y < 0) {
      if (m < 0)
        g = m;
      else
        g = 0;
      j = (g - y) / x;
      k = (g - y) % x;
      count += j + (k == 0 ? 0 : 1);
      y = k == 0 ? g : (g - k + x);
    } else if (x > y) {
      y = x + y;
      count++;
    } else if (y >= x) {
      x = x + y;
      count++;
    }
  }
  cout << count << endl;
}
int main() {
  cin >> a >> b >> c;
  run(a, b, c);
  return 0;
}

```