#include <bits/stdc++.h>
using namespace std;
long long x, y, m, pre;
long long step(long long x, long long y) {
  if (y >= m) return 0;
  if (y <= 0) return -1;
  long long tmp, num = 0;
  if (x) {
    if (x < 0) {
      num = abs(x / y);
      tmp = x + y * num + 2 * y;
      num += 2;
    } else {
      num++;
      tmp = x + y;
    }
  } else {
    num += 2;
    tmp = x + 2 * y;
  }
  x = min(tmp, y);
  y = max(tmp, y);
  return step(x, y) + num;
}
int main(void) {
  while (3 == scanf("%I64d%I64d%I64d", &x, &y, &m)) {
    if (x > y) swap(x, y);
    long long ans = step(x, y);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
long long x, y, m, pre;
long long step(long long x, long long y) {
  if (y >= m) return 0;
  if (y <= 0) return -1;
  long long tmp, num = 0;
  if (x) {
    if (x < 0) {
      num = abs(x / y);
      tmp = x + y * num + 2 * y;
      num += 2;
    } else {
      num++;
      tmp = x + y;
    }
  } else {
    num += 2;
    tmp = x + 2 * y;
  }
  x = min(tmp, y);
  y = max(tmp, y);
  return step(x, y) + num;
}
int main(void) {
  while (3 == scanf("%I64d%I64d%I64d", &x, &y, &m)) {
    if (x > y) swap(x, y);
    long long ans = step(x, y);
    printf("%I64d\n", ans);
  }
  return 0;
}

```