description: For an array c of nonnegative integers, MEX(c) denotes the smallest nonnegative integer that doesn't appear in it. For example, MEX([0, 1, 3]) = 2, MEX([42]) = 0.
  
  You are given integers n, k, and an array [b_1, b_2, …, b_n].
  
  Find the number of arrays [a_1, a_2, …, a_n], for which the following conditions hold:
  
    * 0 ≤ a_i ≤ n for each i for each i from 1 to n.
  
    * |MEX([a_1, a_2, …, a_i]) - b_i| ≤ k for each i from 1 to n.
  
  
  
  
  As this number can be very big, output it modulo 998 244 353.
  
  Input
  
  The first line of the input contains two integers n, k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 50).
  
  The second line of the input contains n integers b_1, b_2, …, b_n (-k ≤ b_i ≤ n+k) — elements of the array b.
  
  Output
  
  Output a single integer — the number of arrays which satisfy the conditions from the statement, modulo 998 244 353.
  
  Examples
  
  Input
  
  
  4 0
  0 0 0 0
  
  
  Output
  
  
  256
  
  
  Input
  
  
  4 1
  0 0 0 0
  
  
  Output
  
  
  431
  
  
  Input
  
  
  4 1
  0 0 1 1
  
  
  Output
  
  
  509
  
  
  Input
  
  
  5 2
  0 0 2 2 0
  
  
  Output
  
  
  6546
  
  
  Input
  
  
  3 2
  -2 0 4
  
  
  Output
  
  
  11
solution: #include <bits/stdc++.h>
  using namespace std;
  template <int MOD_>
  struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");
  
   private:
    using ll = long long;
    int v;
    static int minv(int a, int m) {
      a %= m;
      assert(a);
      return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
    }
  
   public:
    modnum() : v(0) {}
    modnum(ll v_) : v(int(v_ % MOD)) {
      if (v < 0) v += MOD;
    }
    explicit operator int() const { return v; }
    friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
      return out << int(n);
    }
    friend std::istream& operator>>(std::istream& in, modnum& n) {
      ll v_;
      in >> v_;
      n = modnum(v_);
      return in;
    }
    friend bool operator==(const modnum& a, const modnum& b) {
      return a.v == b.v;
    }
    friend bool operator!=(const modnum& a, const modnum& b) {
      return a.v != b.v;
    }
    modnum inv() const {
      modnum res;
      res.v = minv(v, MOD);
      return res;
    }
    friend modnum inv(const modnum& m) { return m.inv(); }
    modnum neg() const {
      modnum res;
      res.v = v ? MOD - v : 0;
      return res;
    }
    friend modnum neg(const modnum& m) { return m.neg(); }
    modnum operator-() const { return neg(); }
    modnum operator+() const { return modnum(*this); }
    modnum& operator++() {
      v++;
      if (v == MOD) v = 0;
      return *this;
    }
    modnum& operator--() {
      if (v == 0) v = MOD;
      v--;
      return *this;
    }
    modnum& operator+=(const modnum& o) {
      v += o.v;
      if (v >= MOD) v -= MOD;
      return *this;
    }
    modnum& operator-=(const modnum& o) {
      v -= o.v;
      if (v < 0) v += MOD;
      return *this;
    }
    modnum& operator*=(const modnum& o) {
      v = int(ll(v) * ll(o.v) % MOD);
      return *this;
    }
    modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
    friend modnum operator++(modnum& a, int) {
      modnum r = a;
      ++a;
      return r;
    }
    friend modnum operator--(modnum& a, int) {
      modnum r = a;
      --a;
      return r;
    }
    friend modnum operator+(const modnum& a, const modnum& b) {
      return modnum(a) += b;
    }
    friend modnum operator-(const modnum& a, const modnum& b) {
      return modnum(a) -= b;
    }
    friend modnum operator*(const modnum& a, const modnum& b) {
      return modnum(a) *= b;
    }
    friend modnum operator/(const modnum& a, const modnum& b) {
      return modnum(a) /= b;
    }
  };
  template <typename T>
  T pow(T a, long long b) {
    assert(b >= 0);
    T r = 1;
    while (b) {
      if (b & 1) r *= a;
      b >>= 1;
      a *= a;
    }
    return r;
  }
  using num = modnum<998244353>;
  vector<num> fact, ifact;
  void init() {
    int N = 1100000;
    fact = {1};
    for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
    ifact.resize(N);
    ifact.back() = 1 / fact.back();
    for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
  }
  num ncr(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    init();
    int n, k;
    cin >> n >> k;
    vector<int> l(n), r(n);
    for (int i = 0; i < n; i++) {
      int b;
      cin >> b;
      l[i] = max(0, b - k);
      r[i] = min(n, b + k);
      if (l[i] > r[i]) {
        cout << 0 << '\n';
        return 0;
      }
    }
    vector<vector<num> > dp(1, vector<num>(1, 1));
    int pl = 0;
    int pr = 0;
    for (int i = 0; i < n; i++) {
      int nl = l[i];
      int nr = r[i];
      vector<vector<num> > ndp(i + 2, vector<num>(nr - nl + 1, 0));
      for (int a = 0; a < dp.size(); a++) {
        for (int pmex = 0; pmex < dp[a].size(); pmex++) {
          dp[a][pmex] *= fact[a];
        }
      }
      for (int T = 0 + pl; T < (int)dp.size() + pr; T++) {
        int L = 0;
        int R = 0;
        num cur = 0;
        for (int nmex = nl; nmex <= nr && nmex <= T + 1; nmex++) {
          if (T - (nmex - 1) >= ndp.size()) continue;
          int r = min(pr, nmex - 1) + 1;
          int l = max(pl, T - int(dp.size()) + 1);
          if (l < r) {
            while (L < l && L < R) {
              cur -= dp[T - L][L - pl];
              L++;
            }
            while (R > r && R > L) {
              R--;
              cur -= dp[T - R][R - pl];
            }
            if (L == R) {
              L = R = l;
            }
            while (R < r) {
              cur += dp[T - R][R - pl];
              R++;
            }
            while (L > l) {
              L--;
              cur += dp[T - L][L - pl];
            }
          }
          ndp[T - (nmex - 1)][nmex - nl] += cur;
        }
      }
      for (int a = 0; a < dp.size(); a++) {
        for (num& x : dp[a]) x *= ifact[a];
      }
      for (int a = 0; a < ndp.size(); a++) {
        for (num& x : ndp[a]) x *= ifact[a];
      }
      for (int a = 0; a < dp.size(); a++) {
        for (int pmex = 0; pmex < dp[a].size(); pmex++) {
          int mex = pl + pmex;
          if (mex >= nl && mex <= nr) {
            ndp[a + 1][mex - nl] += dp[a][pmex] * 1;
            ndp[a][mex - nl] += dp[a][pmex] * (mex + a);
          }
        }
      }
      pl = nl;
      pr = nr;
      dp = ndp;
    }
    num ans = 0;
    for (int a = 0; a < dp.size(); a++) {
      for (int pmex = 0; pmex < dp[a].size(); pmex++) {
        int mex = pl + pmex;
        ans += dp[a][pmex] * ncr(n - mex, a) * fact[a];
      }
    }
    cout << ans << '\n';
  }
  