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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long x = in.nextLong();
            long y = in.nextLong();
            long m = in.nextLong();
            if (Math.max(x, y) >= m) out.println(0);
            else if (x <= 0 && y <= 0) {
                out.println(-1);
            } else {
                long ans = 0;
                if (x < 0) {
                    ans += -x / y;
                    x += -y * (x / y);
                }
                if (x < 0) {
                    x += y;
                    ans++;
                }
                if (y < 0) {
                    ans += -y / x;
                    y += -x * (y / x);
                }
                if (y < 0) {
                    y += x;
                    ans++;
                }
                long dis = m - Math.max(x, y);
                while (dis > 0) {
                    ans++;
                    if (x > y) {
                        y += x;
                    } else {
                        x += y;
                    }
                    dis = m - Math.max(x, y);
                }
                out.println(ans);
            }
        }

    }

    static class InputReader {
        private StringTokenizer tokenizer;
        private BufferedReader reader;

        public InputReader(InputStream inputStream) {
            reader = new BufferedReader(new InputStreamReader(inputStream));
        }

        private void fillTokenizer() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (Exception e) {
                    throw new RuntimeException(e);
                }
            }
        }

        public String next() {
            fillTokenizer();
            return tokenizer.nextToken();
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}

