#include <bits/stdc++.h>
using namespace std;
const long long N = 200;
const double eps = 1e-6;
const double pi = 4 * atan(1);
long long n, t;
long long mx, mn, m;
int solve() {
  cin >> mx >> mn >> m;
  if (mx < mn) swap(mx, mn);
  if (mx >= m)
    printf("0\n");
  else {
    long long k = 0;
    if (mx > 0) k = (mx - mn) / mx;
    mn = (mx * k) + mn;
    for (int i = 1; i <= 300; i++) {
      long long x = mn + mx;
      if (mn > m || mx > m) break;
      mn = max(mn, x);
      if (mx < mn) swap(mn, mx);
      if (mx >= m) {
        printf("%lld\n", i + k);
        return 0;
      }
    }
    printf("-1\n");
  }
  return 0;
}
int main() { return solve(); }

### Prompt

Your task is to create 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;
const long long N = 200;
const double eps = 1e-6;
const double pi = 4 * atan(1);
long long n, t;
long long mx, mn, m;
int solve() {
  cin >> mx >> mn >> m;
  if (mx < mn) swap(mx, mn);
  if (mx >= m)
    printf("0\n");
  else {
    long long k = 0;
    if (mx > 0) k = (mx - mn) / mx;
    mn = (mx * k) + mn;
    for (int i = 1; i <= 300; i++) {
      long long x = mn + mx;
      if (mn > m || mx > m) break;
      mn = max(mn, x);
      if (mx < mn) swap(mn, mx);
      if (mx >= m) {
        printf("%lld\n", i + k);
        return 0;
      }
    }
    printf("-1\n");
  }
  return 0;
}
int main() { return solve(); }

```