Let f(x)=∣2{x}−1∣ where {x} denotes the fractional part of x. The number n is the smallest positive integer such that the equation nf(xf(x))=x has at least 2012 real solutions x. What is n?<spanclass=′latex−bold′>Note:</span> the fractional part of x is a real number y={x}, such that 0≤y<1 and x−y is an integer. <spanclass=′latex−bold′>(A)</span>30<spanclass=′latex−bold′>(B)</span>31<spanclass=′latex−bold′>(C)</span>32<spanclass=′latex−bold′>(D)</span>62<spanclass=′latex−bold′>(E)</span>64