> For the complete documentation index, see [llms.txt](https://solutions.icpc.uclaacm.com/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://solutions.icpc.uclaacm.com/2021-tryout-solutions/tryout-2-solutions/f-base-2-palindromes.md).

# F: Base-2 Palindromes

* The key observation in this problem is that a palindrome is uniquely defined by the first half bits. Thus, the number of palindromes of length $$n$$ ($$n>=2$$) with the first bit $$1$$ is $$2^{\lceil\frac{n-2}{2}\rceil}$$.
* We can figure out the most significant bit of the $$m^{th}$$ palindrome by counting the number of palindromes of each length until the total number of palindromes we have counted so far exceeds $$m$$.
* Suppose the most significant bit of the $$m^{th}$$ palindrome is the $$b^{th}$$ bit, and suppose there are $$k$$ palindromes of length at most $$b-1$$. Then, we are interested in the $$m' = (m-k)^{th}$$​palindrome of length $$b$$ . 
* Let $$s$$ denote the binary representation of $$m-k-1$$. Let $$s\_{rev}$$ denote the string $$s$$ reversed. Now, we know that the answer is $$1ss\_{rev}1$$ (expect if $$b$$ is odd, in which case we remove one bit from $$s\_{rev}$$​). 
  * For instance, suppose $$s=0101$$. If $$b$$ is even, the answer is $$1010110101$$. If b is odd, the answer is $$101010101$$.
* Finally, convert the binary string to an integer.


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