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>
long long max(long long a, long long b) { return a < b ? b : a; }
long long min(long long a, long long b) { return a > b ? b : a; }
int main() {
  long long x, y, m;
  while (scanf("%I64d%I64d%I64d", &x, &y, &m) == 3) {
    if (x <= 0 && y <= 0) {
      if (max(x, y) >= m) {
        printf("0\n");
      } else
        printf("-1\n");
      continue;
    }
    if (max(x, y) >= m) {
      printf("0\n");
      continue;
    }
    long long ans = 0;
    long long t1 = min(x, y);
    long long t2 = max(x, y);
    if (t1 < 0) {
      ans += -t1 / t2;
      if ((-t1) % t2 != 0) ans++;
      t1 = t1 + ans * t2;
    }
    while (t2 < m) {
      x = t2;
      y = t1 + t2;
      t1 = min(x, y);
      t2 = max(x, y);
      ans++;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
