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 BSRK Aditya
 */
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);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        long num1 = Long.parseLong(in.next());
        long num2 = Long.parseLong(in.next());
        long perfect_num = Long.parseLong(in.next());

        if (num1 <= 0 && num2 <= 0 && Math.max(num1, num2) < perfect_num)
            out.println("-1");
        else {
            long num_op = 0;
            while (Math.max(num1, num2) < perfect_num) {
                if (num1 < num2) {
                    long t = num1;
                    num1 = num2;
                    num2 = t;
                }
                long k;
                if (num1 == num2) k = 1;
                else k = Mathy.ceilDiv(num1 - num2, num1);
                num2 += k*num1;
                num_op += k;
            }
            out.println(num_op);
        }
    }
}

class Mathy {

    public static long ceilDiv(long x, long y) {
        return (x/y) + (x%y==0?0:1);
    }
}

