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.math.BigInteger;
import java.util.Scanner;


public class Main {
    static BigInteger maxBi(BigInteger a, BigInteger b){
        if (a.compareTo(b) > 0)
            return a;
        return b;
    }
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        long x = scan.nextLong(), y = scan.nextLong(), m = scan.nextLong();
        BigInteger bx = BigInteger.valueOf(x);
        BigInteger by = BigInteger.valueOf(y);
        BigInteger bm = BigInteger.valueOf(m);
        if (x <= 0 && y <= 0 && Math.max(x, y) < m){
            System.out.println(-1);
            return;
        }
        if (Math.max(x, y) >= m){
            System.out.println(0);
            return;
        }
        BigInteger cnt = BigInteger.ZERO;
        if (bx.compareTo(by) < 0){
            BigInteger t = bx;
            bx = by;
            by = t;
        }
        if (by.compareTo(BigInteger.ZERO) < 0){
            BigInteger t = by.abs();
            cnt = cnt.add(t.divide(bx));
            by = by.add(bx.multiply(cnt));
        }
        while (maxBi(bx, by).compareTo(bm) < 0){
            cnt = cnt.add(BigInteger.ONE);
            BigInteger t = bx.add(by);
            if (bx.compareTo(by) < 0)
                bx = t;
            else
                by = t;
        }
        System.out.println(cnt);
    }
}
