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.
#include <bits/stdc++.h>
using namespace std;
long long x, y, m;
long long F[90];
void init() {
  F[0] = F[1] = 1;
  for (int i = 2; i < 90; ++i) F[i] = F[i - 1] + F[i - 2];
}
int main() {
  init();
  while (cin >> x >> y >> m) {
    if (x > y) swap(x, y);
    if (x >= m || y >= m)
      puts("0");
    else if (x == 0 && y == 0)
      puts("-1");
    else if (x <= 0 && y <= 0 && m > y)
      puts("-1");
    else if (x >= 0 && y >= 0) {
      for (int i = 0; i < (90); i++)
        if (i && F[i] * y + F[i - 1] * x >= m) {
          cout << i << endl;
          break;
        }
    } else if (x <= 0 && y <= 0) {
      for (int i = 0; i < (90); i++)
        if (i && F[i] * x + F[i - 1] * y >= m) {
          cout << i << endl;
          break;
        }
    } else {
      x = -x;
      long long cnt = x / y + ((x % y) > 0);
      x = -x;
      x = x + y * cnt;
      for (int i = 0; i < (90); i++)
        if (i && F[i] * y + F[i - 1] * x >= m) {
          cout << i + cnt << endl;
          break;
        }
    }
  }
}
