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.*;
import java.util.*;

public class Contest188 {

    public static void main(String[] args) throws IOException {
        new Contest188().run();
    }

    void solve() throws IOException {
        long x = nextLong();
        long y = nextLong();
        long m = nextLong();

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

        if(x <= 0 && y <= 0) {
            out.println(-1);
            return;
        }

        long cnt = 0;

        if(y < x) {
            long tmp = x;
            x = y;
            y = tmp;
        }

        if(x < 0) {
            long target = Math.min(m, 0);
            long dist = target - x;
            long steps = (dist + y - 1) / y;
            cnt += steps;
            x += steps * y;
            if(y < x) {
                long tmp = x;
                x = y;
                y = tmp;
            }
        }

        while(y < m) {
            long next = x + y;
            x = y;
            y = next;
            ++cnt;
        }

        out.println(cnt);
    }

    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

    public String nextToken() throws IOException {
        while(st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }

        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws  IOException {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    void run() throws IOException {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;

        br = new BufferedReader( new InputStreamReader( oj ? System.in : new FileInputStream("input.txt")));
        out = new PrintWriter( oj ? System.out : new FileOutputStream("output.txt"));

        solve();

        out.close();
    }

}
