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

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

class TaskA {
  public void solve(int testNumber, FastScanner in, PrintWriter out) {
    long x = in.nextLong();
    long y = in.nextLong();
    long m = in.nextLong();

    if (x >= m || y >= m) {
      out.println(0);
      return;
    }

    if (x <= 0 && y <= 0) {
      out.println(-1);
      return;
    }
    long cnt = 0;
    // one is > 0
    if (x < 0) {
      cnt += (-x / y);
      x += (y * (-x / y));
    }
    if (y < 0) {
      cnt += (-y / x);
      y += (x * (-y / x));
    }

    while (x < m && y < m) {
      if (x < y) {
        x += y;
      } else {
        y += x;
      }
      cnt++;
    }

    out.println(cnt);
  }
}

class FastScanner {
  BufferedReader br;
  StringTokenizer st;

  public FastScanner(InputStream in) {
    br = new BufferedReader(new InputStreamReader(in));
    st = new StringTokenizer("");
  }

  public String next() {
    try {
      if (st.hasMoreTokens()) return st.nextToken();
      st = new StringTokenizer(br.readLine());
      return next();
    } catch (Exception e) {
      return "";
    }
  }

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

