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.
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
//package pkg188;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

/**
 *
 * @author Alik
 */
public class C {
    static String line;
    static int cur = 0;
    static long a, b, m, ans = 0;
    public static void main(String[] args) throws Exception{
        Scanner in = new Scanner(System.in);
        a = in.nextLong(); b = in.nextLong(); m = in.nextLong();
        if (Math.max(a, b) >= m) {
            System.out.println(0);
            return;
        }
        if (a <= 0 && b <= 0) {
            System.out.println("-1");
            return;
        }
        long q = Math.min(a, b);
        if (q >= 0) {find(a, b); return;}
        ans = Math.abs(q) / Math.max(a, b);
        if (a < b) a = a + ans * b; else b = b + ans * a;
        find(a, b);
    }

    private static void find(long a, long b) {
        long cc = 0;
        while (a < m && b < m) {
            if (a < b) a = a + b; else b = a + b;
            cc++;
        }
        System.out.println(ans + cc);
    }
    
   
}
