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;
int main() {
  long long a, b, m, t, x, c = 0, L, q;
  while (cin >> a >> b >> m) {
    t = 0;
    for (; a < m && b < m;) {
      q = 0;
      x = a + b;
      long long p = min(a, b);
      long long k = max(a, b);
      if (m > 0 && x <= p) {
        c = 1;
        cout << "-1" << endl;
        break;
      }
      if (m < 0 && x <= k) {
        c = 1;
        cout << "-1" << endl;
        break;
      }
      if (x < 0) {
        L = abs(p);
        long long d = L / k;
        if (L % k != 0) d++;
        x = k * d - L;
        t += d;
        q = 1;
      }
      if (a < b)
        a = x;
      else
        b = x;
      if (!q) t++;
    }
    if (!c) cout << t << endl;
  }
}
