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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PerfectPair317A {

    public static void main(String[] args) throws IOException {
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tok = new StringTokenizer(buffer.readLine());
        long x = Long.parseLong(tok.nextToken());
        long y = Long.parseLong(tok.nextToken());
        long m = Long.parseLong(tok.nextToken());
        long count = 0;
        long help;
        if(x>y){
            help= x; x = y; y =help;
        }
        if(m <=y){
            System.out.println("0");
            System.exit(0);
        }
        if(1>y){ 
            System.out.println("-1");
            System.exit(0);
        }
        for(count = (y-x)/y, x+=y*count; m>y; ++count){
            help = x+y;
            x = y;
            y = help;
        }
        System.out.println(count);
    }
}
