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.Scanner;
public class Main{
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		long a = sc.nextLong();
		long b = sc.nextLong();
		long m = sc.nextLong();
		long ans = -1;
		if(m<=a || m<=b){
			ans++;}
		else {
			if(a>=0 && b>=0){
				if(a==0 && b==0){}
				else{
					ans++;
					while(Math.max(a,b)<m){
						if(a>=b){
							b=a+b;
							ans++;}
						else{
							a=a+b;
							ans++;}}}}
			else if(Math.min(a,b)<0 && Math.max(a,b)>0){
				ans++;
				long min = Math.min(a,b);
				long max = Math.max(a,b);
				long n = Math.abs(min/max)+1;
				ans = ans+n;
				long x = max-Math.abs(min%max);
				long y = max;
				while(Math.max(x,y)<m){
						if(x>=y){
							y=y+x;
							ans++;}
						else{
							x=x+y;
							ans++;}}}
			else if(a<0 && b<0){}}
	System.out.println(ans);}}
			
			
			
			
			