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;
const double pi = acos(-1);
const double tau = 2 * pi;
const double epsilon = 1e-9;
long long X, Y, M;
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> X >> Y >> M;
  if (X >= M || Y >= M)
    cout << "0\n";
  else if (X + Y <= X && X + Y <= Y)
    cout << "-1\n";
  else {
    long long nOperations = 0;
    if (X < Y) X ^= Y ^= X ^= Y;
    if (Y < 0) {
      nOperations += (-Y / X) + ((-Y) % X != 0);
      if (-Y % X != 0)
        Y = X - (-Y) % X;
      else
        Y = 0;
    }
    if (X < Y) X ^= Y ^= X ^= Y;
    while (X < M) {
      Y = X + Y;
      if (X < Y) X ^= Y ^= X ^= Y;
      ++nOperations;
    }
    cout << nOperations << '\n';
  }
  return 0;
}
