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 sys
st=[0]
st.append(1)
st.append(1)
for i in range(3,101,1):
  tmp=st[i-1]+st[i-2]
  st.append(tmp)
  
inp=raw_input()
lis=inp.split()
x=int(lis[0])
y=int(lis[1])
m=int(lis[2])
if  x > y:
  tmp=x
  x=y
  y=tmp
if m < 0:
  if y>=m:
    print "0"
  else:
    print "-1"
else:
  if y>=m:
    print "0"
  elif y<=0:
    print "-1"
  else:
     add=0
     if x < 0:
        tmp=x*-1
        if tmp%y==0:
	  add=tmp/y
	else:
	  add=tmp/y+1
        x=x+y*add
     for i in range(1,100,1):
       if x*st[i]+y*st[i+1] >=m:
         break

     print i+add
