# Problem E

Arithmetic Decoding

Arithmetic coding is a method to represent a message as a real number $x$ such that $0 \leq x < 1$. We will assume that the message consists only of uppercase ‘A’s and ‘B’s. The two letters have associated probabilities $p_ A$ and $p_ B = 1 - p_ A$ such that $0 < p_ A < 1$.

The current interval $[a,b)$ is initially set to $[0,1)$ and we will update this interval one letter at a time. To encode a letter, the current interval is divided into two subintervals as follows. Let $c = a + p_ A(b-a)$. If the next letter is ‘A’, $[a,c)$ becomes the current interval. Otherwise, the current interval is now $[c,b)$. This process is repeated for each letter in the message. If $[k,\ell )$ is the final interval, the encoded message is chosen to be $k$.

For example, if the original message is “ABAB” and $p_ A = p_ B = 0.5$, the sequence of intervals encountered in the algorithm is

\[ [0,1) \xrightarrow {A} [0, 0.5) \xrightarrow {B} [0.25, 0.5) \xrightarrow {A} [0.25, 0.375) \xrightarrow {B} [0.3125, 0.375). \]The encoded message is therefore 0.3125, or 0.0101 in binary.

Given the length of the message, the probabilities, and the encoded message, determine the original message.

## Input

The first line contains the integer $N$ ($1 \leq N \leq 15$), which is the length of the original message. The second line contains the integer $D$ ($1 \leq D \leq 7$), which indicates that $p_ A = \frac{D}{8}$. The third line contains the binary representation of the encoded message. It is guaranteed that the binary representation of the encoded message starts with “0.” and contains at most $3N+2$ characters.

It is guaranteed that the encoded message came from an initial message of length $N$ consisting only of ‘A’ and ‘B’ using this value of $p_ A$.

## Output

Display the original message.

Sample Input 1 | Sample Output 1 |
---|---|

4 4 0.0101 |
ABAB |

Sample Input 2 | Sample Output 2 |
---|---|

6 5 0.100100100100101 |
ABBABA |