#include<stdio.h>
#include<vector>
#include<atcoder/modint>
using namespace std;
using namespace atcoder;
using mint=modint998244353;

int mem=0;
vector<vector<mint>> Q_data(60);

mint Bostan_Mori(long long N){
	int d=6;
	vector<mint> Q(d+1);
	if(mem>0){
		Q=Q_data[0];
	}
	if(mem==0){
		Q[0]=1;
		for(int i=1;i<=d;i++){
			Q[i]=-1;
		}
		Q_data[0]=Q;
		mem++;
	}
	vector<mint> P(d);
	P[0]=1;
	int count=0;
	while(N>0){
		count++;
		if(mem<=count){
			vector<mint> nQ(d+1);
			for(int i=0;i<=d;i++){
				for(int j=max(0,2*i-d);j<=min(2*i,d);j++){
					if(j%2==0){
						nQ[i]+=Q[j]*Q[2*i-j];
						continue;
					}
					nQ[i]-=Q[j]*Q[2*i-j];
				}
			}
			Q_data[mem]=nQ;
			mem++;
		}
		vector<mint> nP(d);
		if(N%2==0){
			for(int i=0;i<2*d;i+=2){
				for(int j=max(0,i-d+1);j<=min(d,i);j++){
					if(j%2==0){
						nP[i/2]+=Q[j]*P[i-j];
						continue;
					}
					nP[i/2]-=Q[j]*P[i-j];
				}
			}
			P=nP;
		}
		if(N%2==1){
			for(int i=1;i<2*d;i+=2){
				for(int j=max(0,i-d+1);j<=min(d,i);j++){
					if(j%2==0){
						nP[i/2]+=Q[j]*P[i-j];
						continue;
					}
					nP[i/2]-=Q[j]*P[i-j];
				}
			}
			P=nP;
		}
		Q=Q_data[count];
		N/=2;
		continue;
	}
	return P[0];
}

template<typename T>
struct matrix{
	int H,W;
	vector<vector<T>> table;
	matrix(int h,int w) : H(h),W(w){
		table.resize(h,vector<T>(w));
	}
	int size(){
		return H;
	}
	matrix pow(long long N){
		assert(H==W);
		assert(0<=N);
		matrix x=*this;
		matrix r(H,H);
		for(int i=0;i<H;i++){
			r[i][i]=1;
		}
		while(N){
			if(N&1)r*=x;
			x*=x;
			N>>=1;
		}
		return r;
	}
	vector<T>& operator[](int index){
		assert(index<H);
		return table[index];
	}
	matrix& operator+=(const matrix &other){
		for(int i=0;i<H;i++){
			for(int j=0;j<W;j++){
				if((int)(other.table.size())<=i){
					continue;
				}
				if((int)(other.table[i].size())<=j){
					continue;
				}
				(*this)[i][j]+=other.table[i][j];
			}
		}
		return *this;
	}
	matrix operator*=(const matrix &other){
		int h=other.H;
		int w=other.W;
		assert(h==W);
		//結果はH*w行列になる
		matrix result(H,w);
		for(int i=0;i<H;i++){
			for(int k=0;k<W;k++){
				for(int j=0;j<w;j++){
					result.table[i][j]+=table[i][k]*other.table[k][j];
				}
			}
		}
		*this=result;
		return *this;
	}
};

int main(){
	int N,k;
	long long M;
	scanf("%d%lld%d",&N,&M,&k);
	//1,1,2,4,8,16で始まり，前6項を足す線型漸化式に使うvectorを用意しておく
	vector<mint> A(6);
	A[0]=1;
	A[1]=1;
	A[2]=2;
	A[3]=4;
	A[4]=8;
	A[5]=16;
	vector<mint> c(6,mint(1));
	for(;k--;){
		int C;
		scanf("%d",&C);
		vector<vector<mint>> P(6,vector<mint>(6));
		//P[i][j]:X=iからスタートして，X=N+j(ピッタリ)で終わるサイコロの振り方であって，常にX \not \equiv C (mod N)である場合
		for(int i=0;i<6;i++){
			for(int j=0;j<6;j++){
				if(i==C){
					P[i][j]=0;
					continue;
				}
				if(j==C){
					P[i][j]=0;
					continue;
				}
				if(i<C && j<C){
					P[i][j]=Bostan_Mori(C-i);
					P[i][j]*=Bostan_Mori(N+j-C);
					P[i][j]=Bostan_Mori(N+j-i)-P[i][j];
					continue;
				}
				if(C<j && C<i){
					P[i][j]=Bostan_Mori(N+C-i);
					P[i][j]*=Bostan_Mori(j-C);
					P[i][j]=Bostan_Mori(N+j-i)-P[i][j];
					continue;
				}
				if(i<C && C<j){
					mint res1=Bostan_Mori(C-i);
					mint res2=Bostan_Mori(N);
					mint res3=Bostan_Mori(j-C);
					mint res4=Bostan_Mori(N+C-i);
					mint res5=Bostan_Mori(N+j-C);
					mint res6=Bostan_Mori(N+j-i);
					P[i][j]=res6-res1*res5-res4*res3+res1*res2*res3;
					continue;
				}
				P[i][j]=Bostan_Mori(N+j-i);
			}
		}
		matrix<mint> Q(6,6);
		for(int i=0;i<6;i++){
			for(int j=0;j<6;j++){
				Q[i][j]=P[j][i];
			}
		}
		Q=Q.pow(M-2);//to do:M=1の場合に場合分けして書く
		matrix<mint> R(6,1);
		for(int i=0;i<6;i++){
			R[i][0]=P[0][i];
		}
		Q*=R;
		vector<mint> S(6);
		for(int i=0;i<6;i++){
			if(Q[i][0]==0){
				continue;
			}
			for(int j=0;j<6;j++){
				//iからN+jに動く
				//N~N+jの間のCについては無視
				if(i<C){
					mint res=Bostan_Mori(C-i)*Bostan_Mori(N+j-C);
					res=Bostan_Mori(N+j-i)-res;
					S[j]+=res*Q[i][0];
					continue;
				}
				assert(i!=C);
				mint res=Bostan_Mori(N+j-i);
				S[j]+=res*Q[i][0];
			}
		}
		mint ans=0;
		for(int i=5;i>=0;i--){
			for(int j=i-1;j>=0;j--){
				S[i]-=S[j];
			}
			ans+=S[i];
		}
		printf("%d\n",ans.val());
	}
}