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.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class A {
    static Scanner in; static long next() throws Exception {return in.nextLong();};
//	static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;}
//	static BufferedReader in;
    static PrintWriter out;

    public static void main(String[] args) throws Exception {
        in = new Scanner(System.in);
//		in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
//		in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);

        long a = next(), b = next();
        long c = next();

        long time = 0;
        if (a < c && b < c && a <= 0&& b <= 0) out.println(-1);
        else {
            if (a > b) {
                long t = a;
                a = b;
                b = t;
            }
            if (a < 0 && b < c) {
                time += (-a)/b;
                a += ((-a)/b)*b;
            }
            while (b < c)  {
                a += b;
                if (a > b) {
                    long t = a;
                    a = b;
                    b = t;
                }
                time++;
            }
            out.println(time);
        }
        out.println();
        out.close();
    }
}