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?
Single line of the input contains three integers *x*, *y* and *m* ( - 10<sup class="upper-index">18</sup> ≤ *x*, *y*, *m* ≤ 10<sup class="upper-index">18</sup>).

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.
a,b,m = map(int,input().split())
y=a+b
prev=y-1
cnt = 0
while  y>prev and a<m and b<m:
   if a<b:
      a+=b
   else:
      b+=a
   cnt+=1
   prev=y
   y = a+b
   # print(a,b,y)
if a>=m or b>=m:
   print(cnt)
else:
   print(-1)
   