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 mx = 1000001;
const int mod = (int)1073741824;
int aa[1000011] = {};
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  if (a <= 0 && b <= 0) {
    if (a == 0 && b == 0) {
      if (c <= 0)
        cout << 0;
      else
        cout << -1;
    } else if (c > a && c > b) {
      cout << -1;
    } else if (c <= a || c <= b)
      cout << 0;
    else {
      long long cnt = 0;
      long long mn = min(a, b);
      while (mn > c) {
        if (a < b) {
          b = b + a;
          mn = b;
        } else {
          a = a + b;
          mn = a;
        }
        cnt++;
      }
      cout << cnt;
    }
  } else if (a >= 0 && b >= 0) {
    if (a == 0 && b == 0) {
      if (c <= 0)
        cout << 0;
      else
        cout << -1;
    } else if (c < a && c < b) {
      cout << 0;
    } else {
      long long cnt = 0;
      long long mx = max(a, b);
      while (mx < c) {
        if (a < b) {
          a = a + b;
          mx = a;
        } else {
          b = a + b;
          mx = b;
        }
        cnt++;
      }
      cout << cnt;
    }
  } else if ((a > 0 && b < 0) || (a < 0 && b > 0)) {
    if (c > 0) {
      long long p = max(a, b);
      if (c <= p) {
        cout << 0;
        return 0;
      }
      long long cnt = 0;
      if (a < b) {
        long long gap = abs(a);
        if (gap % b == 0) {
          a = 0;
          cnt = cnt + (gap / b);
        } else {
          cnt = cnt + (gap / b) + 1;
          a = a + gap + b - (gap % b);
        }
        long long mx = max(a, b);
        while (mx < c) {
          if (a < b) {
            a = a + b;
            mx = a;
          } else {
            b = a + b;
            mx = b;
          }
          cnt++;
        }
        cout << cnt;
      } else {
        long long temp = a;
        a = b;
        b = temp;
        long long gap = abs(a);
        if (gap % b == 0) {
          a = 0;
          cnt = cnt + (gap / b);
        } else {
          cnt = cnt + (gap / b) + 1;
          a = a + gap + b - (gap % b);
        }
        long long mx = max(a, b);
        while (mx < c) {
          if (a < b) {
            a = a + b;
            mx = a;
          } else {
            b = a + b;
            mx = b;
          }
          cnt++;
        }
        cout << cnt;
      }
    } else {
      cout << 0;
    }
  }
  return 0;
}
