 0  1  2  3   4         5         6        7  8  9 10  11 12 13 14  15       16     
     0           1        2         3          4            5          6
xx xx xx xx  xxxxxxxx  xxxxxxxx  xxxxxxxx xx xx xx xx  xx xx xx xx  xxxxxxxx
 0  1  2  3   0         0         0        0  1  2  3   0  1  2  3   0        0


Crossover 0
  All of individual 2

Crossover 16
  All of individual 1

Which would mean that we have two new individuals the same as the old. Thus, we
may as well pick a random point between [2,15]. The first byte is always ff.

Say we get a crossover of 2. It would cross at the following. 

 0  1  2  3   4         5         6        7  8  9 10  11 12 13 14  15       16     
     0           1        2         3          4            5          6
xx xx 
 0  1  

       xx xx  xxxxxxxx  xxxxxxxx  xxxxxxxx xx xx xx xx  xx xx xx xx  xxxxxxxx
       2  3   0         0         0        0  1  2  3   0  1  2  3   0        0
