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.util.*;
import java.io.*;
public class Problem {
	static boolean [] damaged;
	public static void main(String[] args)throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		StringTokenizer s = new StringTokenizer(br.readLine());
		long a = Long.parseLong(s.nextToken());
		long b = Long.parseLong(s.nextToken());
		long c = Long.parseLong(s.nextToken());
		out.println(findMin(a, b, c));
		out.close();
		out.close();
	}
	public static long findMin(long a, long b, long c){
		long total = 0;
		// If both are less than or equal to zero, no solution
		if(a <= 0 && b <= 0 && c > a && c > b){
			return -1;
		}
		else{
			while(a < c && b < c){
//				System.out.println("Current total: " + total);
				long operations = beforeSwitch(a, b);
				if(operations == 0){
					total ++;
					if(a > b){
						b = a + b;
					}
					else{
						a = a + b;
					}
				}
				else{
					total += operations;
					if(a > b){
						b += operations * a;
					}
					else{
						a += operations * b;
					}
				}
			}
		}
		return total;
	}
	// Finds total number of times before switch occurs
	public static long beforeSwitch(long a, long b){
		// 1, -2
		return (Math.max(a, b) - Math.min(a,  b)) / Math.max(a, b);
	}
}