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.
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Sanchit M. Bhatnagar (sanchitbhatnagar@gmail.com)
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskC {
  public void solve(int testNumber, Scanner in, PrintWriter out) {
    long X = in.nextLong();
    long Y = in.nextLong();
    long M = in.nextLong();
    if (X >= M || Y >= M) {
      out.println(0);
    } else if (X <= 0 && Y <= 0) {
      out.println(-1);
    } else {
      long count = 0;
      long tmp = X;
      X = Math.min(X, Y);
      Y = Math.max(tmp, Y);
      if (X < 0 && Y > 0) {
        count += (Math.abs(X) + Y - 1) / Y;
        X = X + Y * count;
      }
      while (X < M && Y < M) {
        tmp = X;
        X = Math.min(X, Y);
        Y = Math.max(tmp, Y);
        X = (X + Y);
        count++;
      }
      out.println(count);
    }
  }
}
