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.util.*;
public class codeforce{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong();
        long y = sc.nextLong();
        long m = sc.nextLong();
        long p=0;
        boolean ok = true;
        if(x<=0||y<=0){
            ok = false;
            long hh;
            if(m<=x||m<=y) System.out.println("0");
            else if(x<=0&&y<=0) System.out.println("-1");
            else if(x<=0){
                hh=(1-x)/y;
                if((1-x)%y!=0) hh++;
                x+=hh*y;
                p+=hh;
                ok = true;
            }else{
                hh=(1-y)/x;
                if((1-y)%x!=0) hh++;
                y+=hh*x;
                p+=hh;
                ok = true;
            }
        }
        if(ok){
            //na3taber x et y positif
            long[][] t = new long[10000][2];
            t[1][0] = 1;
            t[1][1] = 1;
            t[2][0] = 1;
            int r=0;
            if(m<=x||m<=y) System.out.println(p);
            else if(m<=x+y) System.out.println((p+1));
            else{
                for(int i=2;i<10000;i++){
                    t[i][1] = t[i][0]+t[i-1][0];
                    t[i+1][0] = t[i][1];
                    if(m<=Math.min(x,y)*t[i][0]+Math.max(x,y)*t[i][1]){
                        r = i;
                        break;
                    }
                }
                System.out.println((r+p));
            }
        }
    }
}