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>
int main() {
  long long int i, arr[10000];
  arr[0] = 0;
  arr[1] = 1;
  arr[2] = 1;
  for (i = 3;; i++) {
    arr[i] = arr[i - 1] + arr[i - 2];
    if (arr[i] > 1000000000000000000) break;
  }
  long long int x, y, m;
  scanf("%lld%lld%lld", &x, &y, &m);
  if (x <= 0 && y <= 0) {
    if (x < m && y < m)
      printf("-1");
    else
      printf("0");
  } else {
    if (x >= m || y >= m) {
      printf("0\n");
      return 0;
    } else {
      if (y < x) {
        long long int temp;
        temp = x;
        x = y;
        y = temp;
      }
      long long int p = 0;
      if (x < 0) {
        p = (long long int)ceil(-1.0 * x / y);
        x = x + p * y;
      }
      if (x >= m || y >= m) {
        printf("%lld\n", p);
        return 0;
      } else {
        long long int la = x + y, lb = y, pa = p + 1;
        if (la >= m || lb >= m) {
          printf("%lld\n", pa);
          return 0;
        } else {
          for (i = 2;; i++) {
            if ((x * arr[i] + y * arr[i + 1]) >= m ||
                (x * arr[i - 1] + y * arr[i]) >= m)
              break;
            pa++;
          }
          la = x + y, lb = x;
          long long int pb = p + 1;
          if (la >= m || lb >= m) {
            printf("%lld\n", pb);
            return 0;
          }
          for (i = 2;; i++) {
            if ((y * arr[i] + x * arr[i + 1]) >= m ||
                (y * arr[i - 1] + x * arr[i]) >= m)
              break;
            pb++;
          }
          printf("%llu\n", (pa > pb ? pb : pa) + 1);
        }
      }
    }
  }
  return 0;
}
