#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include template inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } constexpr long long MAX = 5100000; constexpr long long INF = 1LL << 60; constexpr int inf = 1000000007; //constexpr long long mod = 1000000007LL; constexpr long long mod = 998244353LL; const long double PI = acos((long double)(-1)); using namespace std; typedef unsigned long long ull; typedef long long ll; struct mint { long long x; mint(long long x = 0) :x((x% mod + mod) % mod) {} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod - a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; } mint operator+(const mint a) const { mint res(*this); return res += a; } mint operator-(const mint a) const { mint res(*this); return res -= a; } mint operator*(const mint a) const { mint res(*this); return res *= a; } mint pow(ll t) const { if (!t) return 1; mint a = pow(t >> 1); a *= a; if (t & 1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod - 2); } mint& operator/=(const mint a) { return (*this) *= a.inv(); } mint operator/(const mint a) const { mint res(*this); return res /= a; } }; int main() { /* cin.tie(nullptr); ios::sync_with_stdio(false); */ ll n, q; scanf("%lld %lld", &n, &q); vector a(n); for (int i = 0; i < n; i++) scanf("%lld", &a[i]); vector b(q); for (int i = 0; i < q; i++) scanf("%lld", &b[i]); vector inv_a(n); for (int i = 0; i < n; i++) inv_a[i] = mint(a[i]).inv(); vector> dp(n + 1, vector(n + 1)); dp[0][0] = 1; mint all = 1; for (int i = 0; i < n; i++) { all *= a[i]; for (int j = 0; j <= n; j++) { if (dp[i][j].x == 0) continue; dp[i + 1][j + 1] += dp[i][j] * inv_a[i]; dp[i + 1][j] += dp[i][j] * (mint(1) - inv_a[i]); } } for (int i = 0; i < q; i++) { printf("%lld\n", (dp[n][b[i]] * all).x); } return 0; }