#include #define rep(i,n) for(int i=0;i<(n);i++) using namespace std; class mint{ static const int MOD=998244353; int x; public: mint():x(0){} mint(long long y){ x=y%MOD; if(x<0) x+=MOD; } mint& operator+=(const mint& m){ x+=m.x; if(x>=MOD) x-=MOD; return *this; } mint& operator-=(const mint& m){ x-=m.x; if(x< 0) x+=MOD; return *this; } mint& operator*=(const mint& m){ x=1LL*x*m.x%MOD; return *this; } mint& operator/=(const mint& m){ return *this*=inverse(m); } mint operator+(const mint& m)const{ return mint(*this)+=m; } mint operator-(const mint& m)const{ return mint(*this)-=m; } mint operator*(const mint& m)const{ return mint(*this)*=m; } mint operator/(const mint& m)const{ return mint(*this)/=m; } mint operator-()const{ return mint(-x); } friend mint inverse(const mint& m){ int a=m.x,b=MOD,u=1,v=0; while(b>0){ int t=a/b; a-=t*b; swap(a,b); u-=t*v; swap(u,v); } return u; } friend istream& operator>>(istream& is,mint& m){ long long t; is>>t; m=mint(t); return is; } friend ostream& operator<<(ostream& os,const mint& m){ return os< a(n); rep(i,n) scanf("%d",&a[i]); static mint dp[6001][6001]; // dp[i+1][j] = (a[i] まで見て, 1 を j 個選んだときの場合の数) dp[0][0]=1; rep(i,n) rep(j,i+1) { dp[i+1][j+1]+=dp[i][j]; dp[i+1][j]+=dp[i][j]*(a[i]-1); } rep(_,q){ int b; scanf("%d",&b); printf("%d\n",dp[n][b].to_int()); } return 0; }