A tourist is learning an incorrect way to sort a permutation $(p_1,\ \cdots,\ p_n)$ of the integers $(1,\ \cdots,\ n)$. We define a fix on two adjacent elements $p_i$ and $p_{i+1}$, to be an operation which swaps the two elements if $p_i > p_{i+1}$, and does nothing otherwise. The tourist performs $n-1$ rounds of fixes, numbered $a = 1$, $2$, $\cdots$, $n − 1$. In round a of fixes, the tourist fixes $p_a$ and $p_{a+1}$, then $p_{a+1}$ and $p_{a+2}$, and so on, up to $p_{n−1}$ and $p_n$. In this process, there are $(n − 1) + (n − 2) + · · · + 1 = n(n−1) 2$ total fixes performed. How many permutations of $(1,\ \cdots ,\ 2018)$ can the tourist start with to obtain $(1,\ \cdots ,\ 2018)$ after performing these steps?