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.
x,y,m = map(int,raw_input().split())


def Min(x,y):
 return x<y

if( x >= m or y >= m):
 print 0
 exit()
 
if (not Min(x,y)):
 x,y=y,x
i=0
if( x<0 and y>0):
 x= -x;
 
 i+=(x/y)
 if(x%y!=0):
  i+=1
 x = y*i - x;
 
 #print x,y,i

if( x >= m or y >= m):
 print i
 exit()

nub=0
while(nub<10000):
 if(Min(x,y)):
  x,y= x+y , y
 else:
  x,y= x, x+y  
 i+=1
 nub+=1
 #print x,y,i
 if( x >= m or y >= m):
  print i
  exit()
  
 
 






print -1
