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 int a, b, k;
  cin >> a >> b >> k;
  if (((a < 0 && b < 0) && (k > 0)) || (a < k && b < k && a < 0 && b < 0) ||
      ((a == 0 && b == 0) && k > 0) || (a == 0 && b > 0 && k < 0) ||
      (a == 0 && b < 0 && k > 0) || (b == 0 && a > 0 && k < 0) ||
      (b == 0 && a < 0 && k > 0)) {
    cout << -1 << endl;
  } else if (((a > 0 && b > 0) && (k < 0)) || (a == 0 && b == 0 && k <= 0) ||
             (a >= k || b >= k)) {
    cout << 0 << endl;
  } else {
    long long int ma = max(a, b), count = 0;
    long long int mi = min(a, b);
    long long int next1 = k - 1;
    long long int next2 = k - 1;
    if (a * b < 0 && k > 0) {
      count += (ceil((double)abs(mi) / (double)abs(ma)));
      mi = mi + count * ma;
    } else if (a * b > 0 && k < 0) {
      count += ceil((double)mi / (double)ma);
      ma = ma - count * mi;
    }
    long long int x[2][2], y[2][2], temp, temp2;
    x[0][0] = 1;
    x[0][1] = 1;
    x[1][0] = 1;
    x[1][1] = 0;
    y[0][0] = mi + ma;
    y[0][1] = ma;
    y[1][0] = ma;
    y[1][1] = mi;
    long long int t = 1;
    while (1) {
      if (y[1][1] >= k) {
        break;
      } else if (y[0][1] >= k) {
        break;
      } else if (y[0][0] >= k) {
        count += t;
        break;
      }
      temp = y[0][0];
      y[0][0] = y[0][0] * x[0][0] + y[0][1] * x[1][0];
      temp2 = y[0][1];
      y[0][1] = temp * x[0][1] + y[0][1] * x[1][1];
      y[1][0] = y[0][1];
      y[1][1] = temp2 * x[0][1] + y[1][1] * x[1][1];
      t++;
    }
    cout << count << endl;
  }
  return 0;
}
