Probability that the knight stays on the board

One knight stands on an $N \times N$ chessboard. On every turn the knight picks one of eight directions, each with probability $1/8$, and moves there.

Row 1 is the top row and row $N$ is the bottom row. Column 1 is the leftmost column and column $N$ is the rightmost column. The coordinate $(x, y)$ means row $x$, column $y$.

From $(x, y)$ the knight can move to $(x+1, y+2)$, $(x+2, y+1)$, $(x+2, y-1)$, $(x+1, y-2)$, $(x-1, y-2)$, $(x-2, y-1)$, $(x-2, y+1)$, $(x-1, y+2)$. A direction that leads off the board is picked with the same probability as any other.

Once the knight steps off the board it stops there and never gets back in. Given the starting square and the number of moves $K$, write a program that computes the probability that the knight is still on the board after $K$ moves.

The first line contains $N$, the starting coordinates $x$ and $y$ of the knight, and the number of moves $K$, separated by spaces. ($1 \le N \le 50$, $1 \le x, y \le N$, $0 \le K \le 50$)

Print on the first line the probability that the knight is on the board after $K$ moves, with exactly 10 digits after the decimal point. The answer is a rational number whose denominator is $8^K$, so round that exact value at the 11th digit after the decimal point, rounding a half up, and print all 10 digits. A probability of 1 prints as 1.0000000000 and a probability of 0 prints as 0.0000000000.