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>
template <typename T, typename U>
inline void swap(T &a, U &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
inline void swap(int &a, int &b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
inline void smax(int &a, int b) {
  if (b > a) a = b;
}
inline void smin(int &a, int b) {
  if (b < a) a = b;
}
inline int gcd(int a, int b) {
  if (a < b) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nchose(int a, int b) {
  long long ans = 1;
  smax(a, b - a);
  int p = 2;
  for (int i = (b - a + 1); i <= (b); i++) {
    ans *= (long long)i;
    while (p <= a && ans % p == 0) ans /= p++;
  }
  return ans;
}
using namespace std;
long long a, b, m, res;
int main() {
  cin >> a >> b >> m;
  if (a < b) swap(a, b);
  if (a < m && a < 1) {
    cout << -1;
    return 0;
  }
  while (a < m) {
    if (b < 0) {
      res -= b / a;
      b -= b / a * a;
    }
    res++;
    b = a + b;
    if (b > a) swap(a, b);
  }
  cout << res;
}
