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.
//package codeforces.div2;

import java.util.Scanner;

public class PerfectPair {

    static public void print(long x, long y, long m) {

        long count = 0;
        if(x > y) {
            long p = y;
            y = x;
            x = p;
        }
        if(x>=m || y>=m) {
            System.out.println(0);
        } else {
            boolean flsg = false;
            if (x<0 && y>0){
                count+=Math.abs(x/y);
                x=x%y;
            }else if (x>0 && y<0){
                count+=Math.abs(y/x);
                y=y%x;
            }
            while (y < m) {
                long diff = x + y;
                if (x <= 0 && y <= 0) {
                    flsg = true;
                    break;
                } else {
                    if(x <= 0 && y <= 0) {
                        flsg = true;
                        break;
                    } else {
                        count++;
                        long p = Math.min(y, diff);
                        y = Math.max(y, diff);
                        x = p;
                    }
                }
            }
            if (flsg == false) {
                System.out.println(count);
            } else {
                System.out.println(-1);
            }
        }
    }

    public static void main(String []args) {

        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong();
        long y = sc.nextLong();
        long z = sc.nextLong();
        print(x, y, z);
    }
}
