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 contest;

import java.io.*;
import java.util.*;

public class lockoutb {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        long a = Long.parseLong(st.nextToken());
        long b = Long.parseLong(st.nextToken());
        long to = Long.parseLong(st.nextToken());
        if (a <= 0 && b <= 0 && to >= 0) {
            if ((a == 0 || b == 0) && to == 0) {
                pw.println(0);
            } else {
                pw.println(-1);
            }
        } else if(to <= 0){
            if(a >= to || b >= to){
                pw.println(0);
            }else{
                pw.println(-1);
            }
        }else{
            long ret = 0;
            while (a < to && b < to) {
                if (a >= b) {
                    if(b < 0){
                        ret += -b/a+1;
                        b += (-b/a + 1)*a;
                    }else{
                        b += a; ret++;
                    }
                } else {
                    if(a < 0){
                        ret += -a/b+1;
                        a += (-a/b + 1)*b;
                    }else{
                        a += b; ret++;
                    }
                }
            }
            pw.println(ret);
        }

        pw.close();
        br.close();
    }
}
