**Definitions**
Let $ n \in \mathbb{Z}^+ $ be the number of contests and friends.
Let $ P, Q \in \mathbb{Z} $ such that $ p = \frac{P}{10^6} $, $ q = \frac{Q}{10^6} $, and $ r = 1 - p - q $.
Let $ A = \{a_1, a_2, \dots, a_n\} \subset \mathbb{Z}^+ $ be the set of distinct desired T-shirt sizes.
For each $ a_i \in A $, define the random variable $ X_i $ indicating whether a T-shirt of size $ a_i $ is received:
$$
X_i = \begin{cases}
1 & \text{if at least one T-shirt of size } a_i \text{ is received}, \\
0 & \text{otherwise}.
\end{cases}
$$
**Constraints**
1. $ 1 \le n \le 2 \cdot 10^5 $
2. $ 0 \le P, Q \le 10^6 $, $ P + Q \le 10^6 $
3. $ 1 \le a_i \le 10^9 $, all $ a_i $ distinct
**Objective**
Compute the expected number of friends who receive a T-shirt:
$$
\mathbb{E}\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \mathbb{P}(X_i = 1)
$$
For each $ a_i $, define the set of sizes that can produce a T-shirt of size $ a_i $:
$$
S_i = \{a_i - 1, a_i, a_i + 1\}
$$
Let $ C_j $ be the event that T-shirt size $ j $ is *requested* (i.e., $ j \in A $).
Then the probability that size $ a_i $ is received is:
$$
\mathbb{P}(X_i = 1) = 1 - \prod_{j \in S_i \cap A} \left(1 - \mathbb{P}(\text{size } j \text{ is chosen and yields } a_i)\right)
$$
But more precisely:
For each requested size $ a_i $, the T-shirt of size $ a_i $ can be obtained from:
- Registering for $ a_i $: yields $ a_i $ with prob $ r $, $ a_i - 1 $ with prob $ p $, $ a_i + 1 $ with prob $ q $
- Registering for $ a_i - 1 $: yields $ a_i $ with prob $ q $
- Registering for $ a_i + 1 $: yields $ a_i $ with prob $ p $
Thus, the probability that **no** T-shirt of size $ a_i $ is received is:
$$
\prod_{\substack{k \in A \\ k \in \{a_i - 1, a_i, a_i + 1\}}} \left(1 - \text{prob that registering for } k \text{ produces } a_i\right)
$$
Let $ N_i = \{ k \in A \mid k \in \{a_i - 1, a_i, a_i + 1\} \} $
Then:
$$
\mathbb{P}(X_i = 0) = \prod_{k \in N_i} \left(1 - \begin{cases}
r & \text{if } k = a_i \\
q & \text{if } k = a_i - 1 \\
p & \text{if } k = a_i + 1
\end{cases}\right)
$$
So:
$$
\mathbb{P}(X_i = 1) = 1 - \prod_{k \in N_i} \left(1 - c_{k,i}\right)
$$
where $ c_{k,i} = \begin{cases}
r & k = a_i \\
q & k = a_i - 1 \\
p & k = a_i + 1
\end{cases} $
Finally, compute:
$$
\mathbb{E} = \sum_{i=1}^n \left(1 - \prod_{k \in N_i} (1 - c_{k,i}) \right)
$$
Output $ \mathbb{E} \mod 998244353 $, with all probabilities represented as modular inverses modulo $ 998244353 $.