#include <iostream>
#include <vector>

#include <atcoder/modint>
using namespace atcoder;

using mint = static_modint<998244353>;

using namespace std;
using ll = long long;

constexpr int iINF = 1'000'000'000;
constexpr ll llINF = 1'000'000'000'000'000'000;

int main () {
    const ll MOD = 998244353;
    int N, M, k; cin >> N >> M >> k;
    vector<int> C(k);
    for (int i = 0; i < k; i++) cin >> C[i];

    vector<mint> dp(M * N + 1);
    // dp[i] := 数列の最後がiになる場合の数
    dp[0] = 1;

    for (int i = 0; i <= M * N; i++) {
        for (int j = 1; j <= 6; j++) {
            if (i + j <= M * N) dp[i + j] += dp[i];
        }
    }

    for (int i = 0; i < k; i++) {
        // {C[i]}を踏む事象
        // {C[i] + N}を踏む事象
        // の和集合が取れればよい。

        mint v1 = 0;
        for (int j = 1; j <= 6; j++) {
            if (0 <= M * N - C[i] - j) v1 += dp[M * N - C[i] - j] * (6 - j + 1);
        }
        v1 *= dp[C[i]];

        mint v2 = 0;
        for (int j = 1; j <= 6; j++) {
            if (0 <= M * N - C[i] - N - j) v2 += dp[M * N - C[i] - N - j] * (6 - j + 1);
        }
        v2 *= dp[C[i] + N];

        mint v3 = 0;
        for (int j = 1; j <= 6; j++) {
            if (0 <= M * N - C[i] - N - j) v3 += dp[M * N - C[i] - N - j] * (6 - j + 1);
        }
        v3 *= dp[N] * dp[C[i]];

        mint v = v1 + v2 - v3;
        cout << v.val() << "\n";
    }
}