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 res;
long long a, b, c;
long long minimo(long long primero, long long segundo, long long base) {
  if (primero >= base || segundo >= base) return 0;
  if (primero + segundo >= base) return 1;
  if (primero <= 0 && segundo <= 0) return -1;
  if (primero <= segundo) return 1 + minimo(primero + segundo, segundo, base);
  return 1 + minimo(primero, primero + segundo, base);
}
long long minimo2(long long primero, long long segundo, long long base) {
  long long veces;
  if (primero > segundo) {
    veces = -1 * segundo / primero;
    return veces + minimo(primero, segundo + veces * primero, base);
  } else {
    veces = -1 * primero / segundo;
    return veces + minimo(primero + veces * segundo, segundo, base);
  }
}
int main() {
  cin >> a >> b >> c;
  if (((a < -1000 && b > 0) || (a > 0 && b < -1000)) && (a < c & b < c))
    res = minimo2(a, b, c);
  else
    res = minimo(a, b, c);
  cout << res;
}
