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 dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  if (a >= n || b >= n)
    puts("0");
  else if (a <= 0 && b <= 0)
    puts("-1");
  else if (a >= 0 && b >= 0) {
    long long c = 0;
    while (a < n && b < n) {
      long long x = min(a, b);
      long long y = max(a, b);
      a = y;
      b = x + y;
      c++;
    }
    cout << c << endl;
  } else {
    long long c = 0;
    long long xx = min(a, b);
    long long yy = max(a, b);
    long long r = xx / yy;
    xx += (-r) * yy;
    c -= r;
    a = xx;
    b = yy;
    while (a < n && b < n) {
      long long x = min(a, b);
      long long y = max(a, b);
      a = y;
      b = x + y;
      c++;
    }
    cout << c << endl;
  }
}
