#include #pragma GCC optimize("O3") using namespace std; using ll = long long; #define rep(i,n) for (int i = 0; i < (n); ++i) #define rep2(i,m,n) for (int i = (m); i < (n); ++i) #define rep3(i,a,b) for (int i = (a); i >= (b); --i) #define all(x) (x).begin(),(x).end() inline int popcount(const int x) { return __builtin_popcount(x);} inline ll popcount(const ll x) { return __builtin_popcountll(x);} template void chmin(T &a, const T &b) noexcept { if (b < a) a = b;} template void chmax(T &a, const T &b) noexcept { if (a < b) a = b;} template void drop(const T &x) { std::cout< void debug_out(const T &x, const Args &... args) { std::cout<= 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 { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= 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 { return mint(*this) /= a;} bool operator==(const mint rhs) const { return x == rhs.x; } bool operator!=(const mint rhs) const { return x != rhs.x; } bool operator<(const mint &a) const{ return x>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} int main() { int n, q; cin >> n >> q; vector a(n); rep(i, n) cin >> a[i]; vector > dp(n+5, vector(n+5, 0)); dp[0][0] = 1; // rep(i, n) { // for (int j = 1; j <= i+1; ++j) { // rep(k, i+1) dp[i+1][j] += dp[k][j-1] * mint(a[i] - 1); // } // } for (int j = 1; j <= n; ++j) { for (int i = j-1; i < n; ++i) { // for (int k = 0; k <= i; ++k) { // dp[i+1][j] += dp[k][j-1] * mint(a[i] - 1); // } if (i == 0) dp[i+1][j] = dp[i][j-1] * mint(a[i] - 1); else { dp[i+1][j] = ((dp[i][j]/mint(a[i-1] - 1)) + dp[i][j-1]) * mint(a[i] - 1); } } } while(q--) { int b; cin >> b; mint ans = 0; rep(i, n+1) ans += dp[i][n-b]; cout << ans << "\n"; } return 0; }