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;
int main() {
  scanf("%I64d%I64d%I64d", &x, &y, &M);
  if (x >= M || y >= M) {
    puts("0");
    return 0;
  }
  if (M > 0) {
    if (x <= 0 && y <= 0) {
      puts("-1");
      return 0;
    }
  } else {
    if (x < M && y < M) {
      puts("-1");
      return 0;
    }
  }
  long long t = 0;
  if (x > y) swap(x, y);
  if (y > 0 && x < 0) {
    long long k = -x / y;
    t += k;
    x += k * y;
    if (x < 0) {
      x += y;
      t++;
    }
  }
  while (1) {
    if (x > y) {
      double tmp = (double)x + (double)y;
      if ((double)tmp >= (double)M) {
        t++;
        break;
      }
      y = x + y;
    } else {
      double tmp = (double)y + (double)x;
      if ((double)tmp >= (double)M) {
        t++;
        break;
      }
      x = x + y;
    }
    t++;
  }
  printf("%I64d\n", t);
  getchar();
  getchar();
  return 0;
}
