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 {

    private long solve(long a, long b, long m) {
        if (a >= m || b >= m) {
            return 0;
        }
        if (m < 0) {
            return -1;
        }
        if (a < 0 && b < 0) {
            return -1;
        }
        long part = 0;
        if (b < 0 || a < 0) {
            if (b < 0) {
                long tmp = a;
                a = b;
                b = tmp;
            }
            if (b == 0) {
                return -1;
            }
            part = (-a + b - 1) / b;
            a += part * b;
        }
        long res1 = tryAdd(a, b, m);
        long res2 = tryAdd(b, a, m);
        if (res1 == -1 && res2 == -1) {
            return -1;
        } else if (res2 == -1) {
            return res2 + part;
        } else if (res1 == -1) {
            return res1 + part;
        } else {
            return Math.min(res1, res2) + part;
        }

    }

    private long tryAdd(long aa, long bb, long mm) {
        BigInteger a = BigInteger.valueOf(aa);
        BigInteger b = BigInteger.valueOf(bb);
        BigInteger m = BigInteger.valueOf(mm);
        for (int i = 0; i < 100; i++) {
            if (m.compareTo(a) <= 0 || m.compareTo(b) <= 0) {
                return i;
            }
            BigInteger tmp = a.add(b);
            a = b;
            b = tmp;
        }
        return -1;
    }

    private void run() throws Exception {
        Scanner sc = new Scanner(System.in);
        long[][] vals = {{1, 2, 5},
                {-1, 4, 15},
                {0, -1, 5},
                {1, 0, 1000000000000000000L},
                {1, -1000000000000000000L, 1000000000000000000L}};
        for (long[] v : vals) {
            //System.out.printf("%d %d %d: %d\n", v[0], v[1], v[2], solve(v[0], v[1], v[2]));
        }

        long res = solve(sc.nextLong(), sc.nextLong(), sc.nextLong());

        System.out.println(res);
    }



    public static void main(String[] args) throws Exception {
        new Main().run();
    }
}
