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 long long inf = (long long)1e+9 + 7ll;
const long long linf = (long long)1e+18 + 7ll;
const long double eps = (long double)1e-9;
const long double pi = acosl((long double)-1.0);
const int alph = 26;
static char buff[(int)2e6 + 17];
const int maxn = (int)1e5 + 17;
long long x, y, m;
int main() {
  scanf("%lld %lld %lld", &x, &y, &m);
  if (x > y) swap(x, y);
  long long d = 0;
  for (int i = 0; i < 1488; ++i) {
    if (y >= m) return printf("%lld\n", d), 0;
    if (y <= 0) return printf("-1\n"), 0;
    d += (y - x) / y + 1;
    x += y * ((y - x) / y + 1);
    if (x > y) swap(x, y);
  }
  printf("-1\n");
  return 0;
}
