#include <bits/stdc++.h>
using namespace std;

long long mod = 998244353;
//入力が必ずmod未満の時に使う.
struct mint{
    long long v = 0;
    mint(){} mint(int a){v = a;} mint(long long a){v = a;} mint(unsigned long long a){v = a;}
    long long val(){return v;}
    void modu(){v %= mod;}
 
    mint repeat2mint(const long long a,long long b){
        mint ret = 1,p = a;
        while(b){
            if(b&1) ret *= p;
            p *= p; b >>= 1;
        }
        return ret;
    };
 
    mint operator+(const mint b){return mint(v)+=b;}
    mint operator-(const mint b){return mint(v)-=b;}
    mint operator*(const mint b){return mint(v)*=b;}
    mint operator/(const mint b){return mint(v)/=b;}
 
    mint operator+=(const mint b){
        v += b.v; if(v >= mod) v -= mod;
        return *this;
    }
    mint operator-=(const mint b){
        v -= b.v; if(v < 0) v += mod; 
        return *this;
    }   
    mint operator*=(const mint b){v = v*b.v%mod; return *this;}
    mint operator/=(mint b){
        if(b == 0) assert(false);
        int left = mod-2;
        while(left){if(left&1) *this *= b; b *= b; left >>= 1;}
        return *this;
    }
 
    mint operator++(int){*this += 1; return *this;}
    mint operator--(int){*this -= 1; return *this;}
    bool operator==(const mint b){return v == b.v;}
    bool operator!=(const mint b){return v != b.v;}
    bool operator>(const mint b){return v > b.v;}
    bool operator>=(const mint b){return v >= b.v;}
    bool operator<(const mint b){return v < b.v;}
    bool operator<=(const mint b){return v <= b.v;}
    
    mint pow(const long long x){return repeat2mint(v,x);}
    mint inv(){return mint(1)/v;}
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N,M,K; cin >> N >> M >> K;
    vector<int> C(K);
    for(auto &c : C) cin >> c;

    vector<mint> dp1(N*M),dp2(N*M+9,1);
    dp1.at(0) = 1;
    for(int i=1; i<N*M; i++){
        for(int k=1; k<=6; k++){
            if(i-k < 0) continue;
            dp1.at(i) += dp1.at(i-k);
        }
    }
    for(int i=N*M-1; i>=0; i--){
        for(int k=1; k<=6; k++) dp2.at(i) += dp2.at(i+k);
        dp2.at(i)--;
    }

    for(auto c : C){
        mint answer = dp1.at(c)*dp2.at(c)+dp1.at(c+N)*dp2.at(c+N);
        answer -= dp1.at(c)*dp1.at(N)*dp2.at(c+N);
        cout << answer.v << "\n";
    }

}