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.*;
import java.lang.*;

public class Rextester{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        br.close();
        long a = Long.parseLong(st.nextToken());
        long b = Long.parseLong(st.nextToken());
        long m = Long.parseLong(st.nextToken());
        long count = 0;
        long y = Math.max(a,b);
        long x = Math.min(a,b);
        while(y<m && x<m){
            long prevX = x;
            if(x<0&&y>0){
                long times = Math.abs(x)/y;
                x = Math.abs(x)%y;
                if(x>0){
                    x=-x;
                }
                count+=times;
            }   
            x = x+y;
            if(x<=prevX){
                System.out.println("-1");
                return;
            }
            long t1 = Math.min(x,y);
            long t2 = Math.max(x,y);
            x = t1;
            y = t2;
            count++;
        }
        System.out.println(count);
    }
}