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 PerfectPair2 {

    public static void main(String[] args) {
        Scanner inp = new Scanner(System.in);
        //Get Me "Accpeted" :)
        long x = inp.nextLong(), y = inp.nextLong(), m = inp.nextLong();
        long ans = 0;
        if (x >= m || y >= m) {
            System.out.println(0);
            return;
        }
        if (x < y) {
            long temp = x;
            x = y;
            y = temp;
        }
        if (x <= 0) {
            System.out.println(-1);
            return;
        }
        if (y < 0) {
            y = -y;
            ans = (long) Math.ceil((double) y / (double) x);
            y = ans * x - y;
        } 
         for (int i = 0; x < m; i++) {
            x+=y;
            y=x-y;
           // System.out.println(x);
            ans++;
        }
        System.out.println(ans);
    }
}
