human
Solve in Python:
Creatnx has $n$ mirrors, numbered from $1$ to $n$. Every day, Creatnx asks exactly one mirror "Am I beautiful?". The $i$-th mirror will tell Creatnx that he is beautiful with probability $\frac{p_i}{100}$ for all $1 \le i \le n$.

Some mirrors are called checkpoints. Initially, only the $1$st mirror is a checkpoint. It remains a checkpoint all the time.

Creatnx asks the mirrors one by one, starting from the $1$-st mirror. Every day, if he asks $i$-th mirror, there are two possibilities:  The $i$-th mirror tells Creatnx that he is beautiful. In this case, if $i = n$ Creatnx will stop and become happy, otherwise he will continue asking the $i+1$-th mirror next day.
 In the other case, Creatnx will feel upset. The next day, Creatnx will start asking from the checkpoint with a maximal number that is less or equal to $i$. 

There are some changes occur over time: some mirrors become new checkpoints and some mirrors are no longer checkpoints. You are given $q$ queries, each query is represented by an integer $u$: If the $u$-th mirror isn't a checkpoint then we set it as a checkpoint. Otherwise, the $u$-th mirror is no longer a checkpoint.

After each query, you need to calculate the expected number of days until Creatnx becomes happy.

Each of this numbers should be found by modulo $998244353$. Formally, let $M = 998244353$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

-----Input-----

The first line contains two integers $n$, $q$ ($2 \leq n, q \le 2 \cdot 10^5$)  — the number of mirrors and queries.

The second line contains $n$ integers: $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq 100$).

Each of $q$ following lines contains a single integer $u$ ($2 \leq u \leq n$) — next query.

-----Output-----

Print $q$ numbers – the answers after each query by modulo...
assistant
#Here is Python program to print $q$ numbers – the answers after each query by modulo.

import sys
readline = sys.stdin.readline
from itertools import accumulate
from collections import Counter
from bisect import bisect as br, bisect_left as bl
class PMS:
    #1-indexed
    def __init__(self, A, B, issum = False):
        #Aに初期状態の要素をすべて入れる,Bは値域のリスト
        self.X, self.comp = self.compress(B)
        self.size = len(self.X)
        self.tree = [0] * (self.size + 1)
        self.p = 2**(self.size.bit_length() - 1)
        self.dep = self.size.bit_length()
        
        CA = Counter(A)
        S = [0] + list(accumulate([CA[self.X[i]] for i in range(self.size)]))
        for i in range(1, 1+self.size):
            self.tree[i] = S[i] - S[i - (i&-i)]
        if issum:
            self.sumtree = [0] * (self.size + 1)
            Ssum = [0] + list(accumulate([CA[self.X[i]]*self.X[i] for i in range(self.size)]))
            for i in range(1, 1+self.size):
                self.sumtree[i] = Ssum[i] - Ssum[i - (i&-i)]
    
    def compress(self, L):
        #座圧
        L2 = list(set(L))
        L2.sort()
        C = {v : k for k, v in enumerate(L2, 1)}
        # 1-indexed
        return L2, C
    
    def leng(self):
        #今入っている個数を取得
        return self.count(self.X[-1])
    
    def count(self, v):
        #v(Bの元)以下の個数を取得
        i = self.comp[v]
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
    
    def less(self, v):
        #v(Bの元である必要はない)未満の個数を取得
        i = bl(self.X, v)
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
    
    def leq(self, v):
        #v(Bの元である必要はない)以下の個数を取得
        i = br(self.X, v)
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, v, x):
        #vをx個入れる,負のxで取り出す,iの個数以上取り出すとエラーを出さずにバグる
        i = self.comp[v]
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

    def get(self, i):
        # i番目の値を取得
        if i <= 0:
            return -1
       ...