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 toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long toInt64(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
double toDouble(string s) {
  double r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
bool isUpperCase(char c) { return c >= 'A' && c <= 'Z'; }
bool isLowerCase(char c) { return c >= 'a' && c <= 'z'; }
bool isLetter(char c) { return c >= 'A' && c <= 'Z' || c >= 'a' && c <= 'z'; }
bool isDigit(char c) { return c >= '0' && c <= '9'; }
char toLowerCase(char c) { return (isUpperCase(c)) ? (c + 32) : c; }
char toUpperCase(char c) { return (isLowerCase(c)) ? (c - 32) : c; }
int main() {
  long long x, y, n;
  cin >> x >> y >> n;
  long long count = 0;
  if (x >= n || y >= n) {
    cout << 0 << endl;
    return 0;
  }
  if (x <= 0 && y <= 0) {
    cout << -1 << endl;
    return 0;
  }
  long long taviz;
  if (y > x) {
    taviz = x;
    x = y;
    y = taviz;
  }
  if (y < 0) {
    count += abs(y / x);
    y += abs(y / x) * x;
  }
  while (max(x, y) < n) {
    if (y < x)
      y += x;
    else
      x += y;
    count++;
  }
  cout << count << endl;
}
