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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class CF_188C {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);

        long x = sc.nextLong();
        long y = sc.nextLong();
        long m = sc.nextLong();

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

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

        long count = 1L;

        while( m - Math.max(x,y)> Math.min(y,x)){

            if(x<=0L){
                count += (1L-x)/y + ((1L-x)<y ? 1L:0L);
                x+= y*(((1L-x)/y) + ((1L-x)<y ? 1L:0L));
                continue;
            } else if (y<=0L){
                count += (1L-y)/x + ((1L-y)<x ? 1L:0L);
                y+= x*(((1L-y)/x) + ((1L-y)<x ? 1L:0L));
                continue;           
            }

            count++;
            if(x < y) {
                x = x+y;
            } else {
                y = x+y;
            }
        }

        System.out.println(count);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        public FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return null;
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

        double nextDouble(){
            return Double.parseDouble(next());
        }
    }

}
