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 $998244353$.


-----Examples-----
Input
2 2
50 50
2
2

Output
4
6

Input
5 5
10 20 30 40 50
2
3
4
5
3

Output
117
665496274
332748143
831870317
499122211



-----Note-----

In the first test after the first query, the first and the second mirrors are checkpoints. Creatnx will ask the first mirror until it will say that he is beautiful, after that he will ask the second mirror until it will say that he is beautiful because the second mirror is a checkpoint. After that, he will become happy. Probabilities that the mirrors will say, that he is beautiful are equal to $\frac{1}{2}$. So, the expected number of days, until one mirror will say, that he is beautiful is equal to $2$ and the answer will be equal to $4 = 2 + 2$.

Difficulty: competition

Starter code:

Solutions:
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
        s = 0
        k = self.p
        for _ in range(self.dep):
            if s + k <= self.size and self.tree[s+k] < i:
                s += k
                i -= self.tree[s]
            k //= 2
        return self.X[s]
    
    def gets(self, v):
        #累積和がv以下となる最大のindexを返す
        v1 = v
        s = 0
        k = self.p
        for _ in range(self.dep):
            if s + k <= self.size and self.sumtree[s+k] < v:
                s += k
                v -= self.sumtree[s]
            k //= 2
        if s == self.size:
            return self.leng()
        return self.count(self.X[s]) + (v1 - self.countsum(self.X[s]))//self.X[s]
    
    def addsum(self, i, x):
        #sumを扱いたいときにaddの代わりに使う
        self.add(i, x)
        x *= i
        i = self.comp[i]
        while i <= self.size:
            self.sumtree[i] += x
            i += i & -i
    
    def countsum(self, v):
        #v(Bの元)以下のsumを取得
        i = self.comp[v]
        s = 0
        while i > 0:
            s += self.sumtree[i]
            i -= i & -i
        return s
    
    def getsum(self, i):
        #i番目までのsumを取得
        x = self.get(i)
        return self.countsum(x) - x*(self.count(x) - i)
    
N, Q = map(int, readline().split())
P = list(map(int, readline().split()))
MOD = 998244353
T = [100*pow(pi, MOD-2, MOD)%MOD for pi in P]

AT = [None]*N
AT[0] = T[0]
for i in range(1, N):
    AT[i] = (AT[i-1]+1)*T[i]%MOD
AM = [None]*N
AMi = [None]*N
AM[0] = T[0]
for i in range(1, N):
    AM[i] = AM[i-1]*T[i]%MOD
AMi[N-1] = pow(AM[N-1], MOD-2, MOD)
for i in range(N-2, -1, -1):
    AMi[i] = AMi[i+1]*T[i+1]%MOD
AT += [0]
AM += [1]
AMi += [1]

Ans = [None]*Q
kk = set([0, N])
PM = PMS([0, N], list(range(N+1)))
ans = AT[N-1]
for qu in range(Q):
    f = int(readline()) - 1
    if f not in kk:
        kk.add(f)
        PM.add(f, 1)
        fidx = PM.count(f)
        fm = PM.get(fidx-1)
        fp = PM.get(fidx+1)
        am = (AT[f-1] - AM[f-1]*AMi[fm-1]*AT[fm-1])%MOD
        ap = (AT[fp-1] - AM[fp-1]*AMi[f-1]*AT[f-1])%MOD
        aa = (AT[fp-1] - AM[fp-1]*AMi[fm-1]*AT[fm-1])%MOD
        ans = (ans - aa + am + ap)%MOD
    else:
        kk.remove(f)
        fidx = PM.count(f)
        fm = PM.get(fidx-1)
        fp = PM.get(fidx+1)
        PM.add(f, -1)
        am = (AT[f-1] - AM[f-1]*AMi[fm-1]*AT[fm-1])%MOD
        ap = (AT[fp-1] - AM[fp-1]*AMi[f-1]*AT[f-1])%MOD
        aa = (AT[fp-1] - AM[fp-1]*AMi[fm-1]*AT[fm-1])%MOD
        ans = (ans + aa - am - ap)%MOD
    Ans[qu] = ans
print('\n'.join(map(str, Ans)))