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.
ans = 0

a,b,t = map(int,raw_input().split(" "))

def gagal():
	print -1
	exit(0)

if (a>=t or b>=t):
	print 0
	exit(0)

if (a==0 and b==0 and t>0): gagal()
if (a==0 and b<0 and t>0): gagal()
if (b==0 and a<0 and t>0): gagal()
if (a<0 and b<0 and t>max(a,b)): gagal()

while (a<t and b<t):
	if (a<b):
		dist = abs(a-b)
		tmp = (dist+b-1)/b
		ans += tmp
		a += tmp*b;
	elif (b<a):
		dist = abs(a-b)
		tmp = (dist+a-1)/a
		ans += tmp
		b += tmp*a;
	else:
		ans += 1
		a += b

print ans