#include <bits/stdc++.h>
int main() {
  long long int a, b, m;
  long long int count;
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &m);
  if (a >= m || b >= m) {
    count = 0;
    printf("%lld\n", count);
    return 0;
  }
  if ((a <= 0 && b <= 0)) {
    count = -1;
    printf("%lld\n", count);
    return 0;
  }
  count = 0;
  if (a < 0) {
    count = (-1 * a) / b;
    a = b * count + a;
  }
  if (b < 0) {
    count = (-1 * b) / a;
    b = a * count + b;
  }
  while (a < m && b < m) {
    if (a >= b) {
      b = a + b;
      count++;
    } else {
      a = a + b;
      count++;
    }
  }
  printf("%lld\n", count);
  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>
int main() {
  long long int a, b, m;
  long long int count;
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &m);
  if (a >= m || b >= m) {
    count = 0;
    printf("%lld\n", count);
    return 0;
  }
  if ((a <= 0 && b <= 0)) {
    count = -1;
    printf("%lld\n", count);
    return 0;
  }
  count = 0;
  if (a < 0) {
    count = (-1 * a) / b;
    a = b * count + a;
  }
  if (b < 0) {
    count = (-1 * b) / a;
    b = a * count + b;
  }
  while (a < m && b < m) {
    if (a >= b) {
      b = a + b;
      count++;
    } else {
      a = a + b;
      count++;
    }
  }
  printf("%lld\n", count);
  return 0;
}

```