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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class Haha{
    
    static class FastScanner{
        BufferedReader s;
        StringTokenizer st;
        public FastScanner() throws IOException{
            s = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer(s.readLine());
        }
        public long nextInt() throws IOException{
            if(st.hasMoreTokens())
                return Long.parseLong(st.nextToken());
            else{
                st = new StringTokenizer(s.readLine());
                return nextInt();
            }
        }
    }
    
    public static long sum(long a,long b){
        return (a+b);
    }
    
    public static void main(String[] args) throws IOException{
        FastScanner s = new FastScanner();
        PrintWriter ww = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        long one = s.nextInt();
        long two = s.nextInt();
        long k = s.nextInt();
        long res = 0;
        long max = Math.max(one,two);
        if(max >= k)
            ww.println(0);
        else if(sum(one,two)<0 && (one == 0 || two == 0) && k>0)
            ww.println(-1);
        else if(sum(one,two)>0 && (one == 0 || two == 0) && k<0) 
            ww.println(-1);
        else if(one < 0 && one < k && two <0 && two < k)
            ww.println(-1);
        else if(one == 0 && two == 0)
            ww.println(-1);
        else{
            if(one < 0 && (two == 1 || two == 2 || two == 3)){
                res += (Math.abs(one/two));
                one = 0;
            }
            else if(two < 0 && (one == 1 || one == 2 || one == 3)){
                res += (Math.abs(two/one));
                two = 0;
            }
       //     System.out.println(res);
            while(true){
                if(one <= two)
                    one = sum(one,two);
                else
                    two = sum(one,two);
                res++;
                long ma = Math.max(one,two);
          //      System.out.println(one+" "+two);
                if(ma >= k)
                    break;
            }
             ww.println(res);
        } 
        ww.close();
    }
}