#include <bits/stdc++.h>
using namespace std;
inline int fi() {
  register int c = getchar();
  int x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
  return x;
}
int main() {
  long long x, y, z;
  cin >> x >> y >> z;
  if (x > y) swap(x, y);
  if (y <= 0) {
    puts(y >= z ? "0" : "-1");
  } else if (y >= z)
    puts("0");
  else {
    long long ans = 0;
    if (x < 0) {
      long long t = (-x / y);
      x = (t * y + x);
      ans = t;
    }
    while (y < z) {
      long long tmp = x + y;
      x = y;
      y = tmp;
      if (x > y) swap(x, y);
      ans++;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
inline int fi() {
  register int c = getchar();
  int x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
  return x;
}
int main() {
  long long x, y, z;
  cin >> x >> y >> z;
  if (x > y) swap(x, y);
  if (y <= 0) {
    puts(y >= z ? "0" : "-1");
  } else if (y >= z)
    puts("0");
  else {
    long long ans = 0;
    if (x < 0) {
      long long t = (-x / y);
      x = (t * y + x);
      ans = t;
    }
    while (y < z) {
      long long tmp = x + y;
      x = y;
      y = tmp;
      if (x > y) swap(x, y);
      ans++;
    }
    cout << ans << endl;
  }
  return 0;
}

```