#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------



int N,K;
int A[202020],B[202020];
const ll mo=998244353;
ll dp[202020];
ll dp2[202020];

ll modpow(ll a, ll n = mo-2) {
	ll r=1;a%=mo;
	while(n) r=r*((n%2)?a:1)%mo,a=a*a%mo,n>>=1;
	return r;
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	
	dp[0]=dp2[0]=1;
	cin>>N>>K;
	FOR(i,N) cin>>A[i];
	FOR(i,N) cin>>B[i];
	ll ret=0;
	
	if(K==N) ret=1;
	
	
	FOR(i,N) {
		j=B[i]*modpow(A[i])%mo;
		x=(mo+dp[i]-((i-(K+1)>=0)?dp[i-(K+1)]:0))*j%mo;
		y=(mo+dp2[i]-((i-K>=0)?dp2[i-K]:0))*j%mo;
		dp[i+1]=(dp[i]+x)%mo;
		dp2[i+1]=(dp2[i]+y)%mo;
		if(i+K>=N-1) ret+=x;
		if(i+K-1>=N-1) ret+=mo-y;
	}
	ret%=mo;
	FOR(i,N) ret=ret*A[i]%mo;
	cout<<ret<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}