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.InputStreamReader;
import java.util.StringTokenizer;

public class PerfectPair {
    
    void run() throws Exception {
        BufferedReader bfd = new BufferedReader(
                new InputStreamReader(System.in));
        StringTokenizer tk = new StringTokenizer(bfd.readLine());
        long x = Long.parseLong(tk.nextToken());
        long y = Long.parseLong(tk.nextToken());
        long m = Long.parseLong(tk.nextToken());
        long cnt = 0;
        long max = Math.max(x, y);
        if(max >= m) {
            System.out.println(0);
        } else if (max <= 0)
            System.out.println(-1);
        else { // can
            long mn = Math.min(x, y);
            long mx = Math.max(x, y);
            long linearDiff = 0;
            if (mn < 0) {
                linearDiff = (mx - mn + 1) / mx;
                mn += mx * linearDiff;
            }
            x = mn; y = mx;
            while (x < m && y < m) {
                mn = Math.min(x, y);
                if (x == mn) {
                    x = x + y;
                } else {
                    y = x + y;
                }
                cnt++;
            }
            System.out.println(cnt + linearDiff);
        }
    }
    
    public static void main(String[] args) throws Exception {
        new PerfectPair().run();
    }

}