Let $\{a_k\}^{2011}_{k=1}$ be the sequence of real numbers defined by $$a_1=0.201, \quad a_2=(0.2011)^{a_1},\quad a_3=(0.20101)^{a_2},\quad a_4=(0.201011)^{a_3},$$ and more generally \[ a_k = \begin{cases}(0.\underbrace{20101\cdots0101}_{k+2 \ \text{digits}})^{a_{k-1}}, &\text {if } k \text { is odd,} \\ (0.\underbrace{20101\cdots01011}_{k+2 \ \text{digits}})^{a_{k-1}}, &\text {if } k \text { is even.}\end{cases} \] Rearranging the numbers in the sequence $\{a_k\}^{2011}_{k=1}$ in decreasing order produces a new sequence $\{b_k\}^{2011}_{k=1}$. What is the sum of all the integers $k$, $1\le k \le 2011$, such that $a_k = b_k$? $ \textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012 $