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.io.IOException;
import java.util.StringTokenizer;

public class PerfectPair {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        
        long X = sc.nextLong();
        long Y = sc.nextLong();
        long M = sc.nextLong();
        
    	if (X <= 0 && X < M && Y <= 0 && Y < M) {
    		System.out.println(-1);
    		return;
    	}
        
        long a = Math.min(X, Y);
        long b = Math.max(X, Y);
        
        long cnt = 0;
        while (b < M) {
        	if (a < 0) {
        		long add = (-a + b - 1) / b;
        		cnt += add;
        		a += b * add;
        	} else {
        		cnt++;
        		a += b;
        	}
        	long na = Math.min(a, b);
        	long nb = Math.max(a, b);
        	a = na;
        	b = nb;
        }
        System.out.println(cnt);
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        
        int[] nextIntArray(int n) {
        	int[] arr = new int[n];
        	for (int i = 0; i < n; i++) {
        		arr[i] = nextInt();
        	}
        	return arr;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            String str = "";
            try { str = br.readLine(); }
            catch (IOException e) { e.printStackTrace(); }
            return str;
        }
    }
}