22. In a Cartesian coordinate system, the four vertices of a square are
$$
(2,2), (-2,2), (-2,-2), (2,-2) \text {. }
$$

A particle starts from $(0,0)$. Every second, it moves to one of the eight adjacent integer points (both coordinates are integers) with equal probability, i.e., it moves from point $(x, y)$ to
$$
\begin{array}{l}
(x, y+1), (x+1, y+1), (x+1, y), \\
(x+1, y-1), (x, y-1), (x-1, y-1), \\
(x-1, y), (x-1, y+1)
\end{array}
$$

with a probability of $\frac{1}{8}$. When it reaches the boundary of the square, it stops moving. Let the probability that it finally lands on one of the square's vertices and not on any of the 12 non-vertex boundary points be $\frac{m}{n}$, where $m$ and $n$ are coprime positive integers. Then the value of $m+n$ is ( ).
(A) 4
(B) 5
(C) 7
(D) 15
(E) 39
22. E.

Let $(1,1), (-1,1), (-1,-1), (1,-1)$ be the "inner corner points", $(1,0), (0,1), (-1,0), (0,-1)$ be the inner edge points, and $(0,0)$ be the center point. Let the probability of the particle reaching an inner corner point be $c$, reaching an inner edge point be $e$, and reaching the center point be $d$.

Starting from the center point, the probability of reaching an inner edge point or an inner corner point is $\frac{4}{8}$, then
$$
d=\frac{1}{2} e+\frac{1}{2} c .
$$

Starting from an inner edge point, the probability of reaching another inner edge point, an inner corner point is $\frac{2}{8}$, reaching the center point is $\frac{1}{8}$, and reaching a non-vertex boundary and stopping is $\frac{3}{8}$. Then
$$
\begin{aligned}
e & =\frac{1}{4} e+\frac{1}{4} c+\frac{1}{8} d+\frac{3}{8} \times 0 \\
& =\frac{1}{4} e+\frac{1}{4} c+\frac{1}{8} d .
\end{aligned}
$$

Starting from an inner corner point, the probability of reaching an inner edge point, the center point is $\frac{2}{8}, \frac{1}{8}$, reaching a non-vertex boundary and stopping is $\frac{4}{8}$, and reaching a vertex is $\frac{1}{8}$. Then
$$
\begin{aligned}
c & =\frac{1}{4} e+\frac{1}{8} d+\frac{1}{2} \times 0+\frac{1}{8} \times 1 \\
& =\frac{1}{4} e+\frac{1}{8} d+\frac{1}{8} .
\end{aligned}
$$

Solving equations (1), (2), and (3) simultaneously, we get
$$
d=\frac{4}{35}, e=\frac{1}{14}, c=\frac{11}{70} \text {. }
$$

Since the initial starting point is $(0,0)$, $d$ is the required probability.
Thus, $m+n=4+35=39$.