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.

def solve(x, y, m):
	res = 0
	while max(x, y) < m:
		x, y = x+y, y
		res += 1
		if max(x, y) >= m:
			return res	
		x, y = x, x+y
		res += 1
	return res

def main():
	x, y, m = map(int, input().split())
	if max(x, y) >= m:
		print(0)
		return 
	if x <= 0 and y <= 0:
		print(-1)
		return 
	if x <= 0:
		x, y = y, x
	cnt = 0
	if y < 0:
		cnt = (abs(y)+x-1) // x
		y += cnt * x
	# while y < 0:
	# 	x, y = x, x+y 
	# 	cnt += 1
	cnt += min(solve(x, y, m), solve(y, x, m))
	print(cnt)
main()
			
 	  	 			  	 			 		 	 			   	 	