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.StringTokenizer;
public class Q4 {
    
    public static void main(String[] args)throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out); 
        StringTokenizer st=new StringTokenizer(br.readLine()," ");
        long x=Long.parseLong(st.nextToken());
        long y=Long.parseLong(st.nextToken());
        long m=Long.parseLong(st.nextToken());
        if(x>=m||y>=m)
        {
            writer.println(0);
            writer.flush();
            System.exit(0);
        }
        if(x<m&&y<m&&x<=0&&y<=0)
        {
            writer.println(-1);
            writer.flush();
            System.exit(0);
        }
        long t1=x;long t2=y;
        x=Math.min(t1,t2);
        y=Math.max(t1,t2);
        long c=0;
        if(x<0&&y>0)
        {
            c=(Math.abs(x))/y+1;
            x=y-(Math.abs(x))%y; 
        }
        
        long max=Math.max(x,y);
        while(max<m)
        {
            c++;
            long s=x+y;
            if(x<y)
             x=s;
            else
             y=s;
            max=Math.max(x,y);
            
        }
        writer.println(c);
        writer.flush();  
    }
}