While Gray codes are great, they are also a bit, well... gray^{1}. In this problem, we look at a much more colorful variant.
For an integer $n \ge 1$ and set of integers $P \subseteq \{ 1, \ldots , n\} $, we say that an ordering $(x_1, \ldots , x_{2^ n})$ of all $n$-bit binary strings is an $n$-bit color code with palette $P$, if for all $1 \le i < 2^ n$, it holds that $d(x_ i, x_{i+1}) \in P$, i.e., the number of bits by which any consecutive pair of strings differ is in $P$.
Note that for some palettes, color codes do not exist. For instance, if $n = 6$ and $P = \{ 6\} $, the second string must be the binary negation of the first one, but then the third string must be the negation of the second one, i.e., equal to the first string.
Given $n$ and $P$, can you construct an $n$-bit color code with palette $P$?
The first line of input consists of two integers $n$ ($1 \le n \le 16$) and $p$ ($1 \le p \le n$). Then follow a line with $p$ distinct integers $s_1, \ldots , s_ p$ ($1 \leq s_ i \leq n$ for each $i$) – the elements of $P$.
If there is an $n$-bit color code with palette $P$, output $2^ n$ lines, containing the elements of such a code, in order. If there are many different codes, any one will be accepted. If no such code exists, output “impossible”.
Sample Input 1 | Sample Output 1 |
---|---|
6 1 6 |
impossible |
Sample Input 2 | Sample Output 2 |
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3 1 1 |
000 001 011 010 110 111 101 100 |
Sample Input 3 | Sample Output 3 |
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4 2 3 2 |
0110 1101 1011 0001 0111 1100 1001 0000 0101 0011 1111 1010 0100 1000 0010 1110 |
Footnotes