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 fib(long long x, long long y, long long m) {
  long long f1 = 1, f2 = 1, f3 = 1, k = 1;
  if (y < 0) {
    k += (-y / x);
    y += (-y / x) * x;
  }
  if (y < 0) {
    k++;
    y += x;
  }
  while ((f1 * x + f2 * y) < m) {
    f3 = f1 + f2;
    f2 = f1;
    f1 = f3;
    k++;
  }
  return k;
}
int main() {
  long long x, y, m, MAX, MIN;
  cin >> x >> y >> m;
  MIN = min(x, y);
  MAX = max(x, y);
  if ((MAX < m) && (MAX <= 0))
    cout << "-1";
  else if (MAX >= m)
    cout << "0";
  else
    cout << fib(MAX, MIN, m);
  return 0;
}
