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.awt.Checkbox;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.*;
import java.util.*;

public class factor {

	public static void main(String[] args) throws NumberFormatException,
			IOException {
		// BufferedReader k=new BufferedReader(new
		// InputStreamReader(System.in));
		Scanner k = new Scanner(System.in);
        long num1=k.nextLong();
        long num2=k.nextLong();

        long m=k.nextLong();
        long count=0;
        if (num1>=m||num2>=m) {
			System.out.println(0);
			return;
		}
        if (num1<=0&&num2<=0&&m>=0) {
			System.out.println(-1);
			return;
		}
        if (num1>0&&num2>0&&m<=0) {
			System.out.println(0);
			return;
		}
        if (num1==m||num2==m) {
			System.out.println(0);
			return;
		}
        
        long min=Math.min(num2, num1);
        long max=Math.max(num2, num1);
       
        if (min<0&&max<0&&m<0) {
			
        	min*=-1;
			max*=-1;
			m*=-1;
		}
        
		if (min>m&&max>m) {
			System.out.println(-1);
			return;
		}
        
        if (min<0&&max>0&&min*-1>1000000) {
			long h=min*-1/max;
			count=h;
			min+=h*max+min*-1%max;
		}
        while (max<m&&min<m) {
			long h=max+min;
			long min2=max;
			
			
			min=Math.min(min2, h);
			max=Math.max(min2, h);
			count++;
		}
        System.out.println(count);
        
		//System.out.println(count);
	}

}
