First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$.  Hence, $403$ integers need to be removed.  Next, note that the last digit of each of the products below is $6$.  $1\times2\times3\times4\times6\times7\times8\times9$,  $11\times12\times13\times14\times16\times17\times18\times19$,  $\cdots\cdots$  $2001\times2002\times2003\times2004\times2006\times2007\times2008\times2009$,  and the last digit of the product $2011\times2012\times2013\times2014\times2016\times2017\times2018$ is $4$.  Hence, we need to remove one more "$2$"and the last digit of the product will be $2$.  So the answer is $404$. 
The first $2018$ integers ($1$, $2$, $3$, $\cdots$, $2017$, $2018$) are written on the blackboard. What is the minimum number of integers that should be erased from the blackboard, so that the last digit of the product of the remaining integers is $2$? 

A. $$402$$ 
B. $$403$$ 
C. $$404$$ 
D. $$410$$ 
E. None of the above 