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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char* x) { cerr << '\"' << x << '\"'; }
  void __print(const string& x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V>& x) {
    cerr << '{';
    __print(x.first);
    cerr << ", ";
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T& x) {
    int first = 0;
    cerr << '{';
    for (auto& i : x) cerr << (first++ ? ", " : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const int MOD = 998244353;
  const char nl = '\n';
  const int MX = 100001;
  struct mi {
    long long v;
    explicit operator long long() const { return v; }
    mi() { v = 0; }
    mi(long long _v) {
      v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
      if (v < 0) v += MOD;
    }
    friend bool operator==(const mi& a, const mi& b) { return a.v == b.v; }
    friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
    friend bool operator<(const mi& a, const mi& b) { return a.v < b.v; }
    mi& operator+=(const mi& m) {
      if ((v += m.v) >= MOD) v -= MOD;
      return *this;
    }
    mi& operator-=(const mi& m) {
      if ((v -= m.v) < 0) v += MOD;
      return *this;
    }
    mi& operator*=(const mi& m) {
      v = v * m.v % MOD;
      return *this;
    }
    mi& operator/=(const mi& m) { return (*this) *= inv(m); }
    friend mi pow(mi a, long long p) {
      mi ans = 1;
      assert(p >= 0);
      for (; p; p /= 2, a *= a)
        if (p & 1) ans *= a;
      return ans;
    }
    friend mi inv(const mi& a) {
      assert(a.v != 0);
      return pow(a, MOD - 2);
    }
    mi operator-() const { return mi(-v); }
    mi& operator++() { return *this += 1; }
    mi& operator--() { return *this -= 1; }
    mi operator++(int) {
      v++;
      if (v == MOD) v = 0;
      return mi(v);
    }
    mi operator--(int) {
      v--;
      if (v < 0) v = MOD - 1;
      return mi(v);
    }
    friend mi operator+(mi a, const mi& b) { return a += b; }
    friend mi operator-(mi a, const mi& b) { return a -= b; }
    friend mi operator*(mi a, const mi& b) { return a *= b; }
    friend mi operator/(mi a, const mi& b) { return a /= b; }
    friend ostream& operator<<(ostream& os, const mi& m) {
      os << m.v;
      return os;
    }
    friend istream& operator>>(istream& is, mi& m) {
      long long x;
      is >> x;
      m.v = x;
      return is;
    }
  };
  void __print(mi X) { cout << X.v; }
  mi facs[MX];
  mi facInvs[MX];
  mi choose(mi _a, mi _b) {
    long long a = (long long)_a, b = (long long)_b;
    if (b > a) return 0;
    if (a < 0) return 0;
    if (b < 0) return 0;
    mi cur = facs[a];
    cur = cur * facInvs[b];
    cur = cur * facInvs[a - b];
    return cur;
  }
  void initFacs() {
    facs[0] = 1;
    for (int i = 1; i < (MX); i++) {
      facs[i] = (facs[i - 1] * i);
    }
    facInvs[MX - 1] = inv(facs[MX - 1]);
    for (int i = (MX - 1) - 1; i >= 0; i--) {
      facInvs[i] = facInvs[i + 1] * (i + 1);
    }
  }
  void solve() {
    int N, K;
    cin >> N >> K;
    vector<int> A(N + 1);
    for (int i = 1; i < (N + 1); i++) cin >> A[i];
    for (int i = 1; i < (N + 1); i++) A[i] -= K;
    A[0] = 0;
    K = 2 * K + 1;
    mi cur[N + 1][K];
    cur[0][0] = 1;
    mi nxt[N + 1][K];
    for (int i = 0; i < (N); i++) {
      for (int j = 0; j < (N + 1); j++)
        for (int k = 0; k < (K); k++) nxt[j][k] = 0;
      for (int j = 0; j < (N + 1); j++) {
        for (int k = 0; k < (K); k++) {
          int curMex = A[i] + k;
          int ni = curMex - A[i + 1] + 1;
          if (ni >= K) continue;
          mi mul = 1;
          int nj = j;
          if (ni < 0) {
            ni *= -1;
            if (j < ni) continue;
            mul = choose(j, ni) * facs[ni];
            nj -= ni;
            ni = 0;
          }
          nxt[nj][ni] += cur[j][k] * mul;
        }
      }
      for (int j = (N + 1) - 1; j >= 1; j--) {
        for (int k = 0; k < (K - 1); k++) {
          nxt[j - 1][k + 1] += nxt[j][k] * j;
        }
      }
      for (int j = 0; j < (N + 1); j++) {
        for (int k = 0; k < (K); k++) {
          int curMex = A[i] + k;
          int ni = curMex - A[i + 1];
          if (ni < 0 || ni >= K) continue;
          nxt[j][ni] += cur[j][k] * (curMex + j);
          if (j < N) nxt[j + 1][ni] += cur[j][k];
        }
      }
      for (int j = 0; j < (N + 1); j++)
        for (int k = 0; k < (K); k++) cur[j][k] = nxt[j][k];
    }
    mi ans = 0;
    for (int i = 0; i < (N + 1); i++) {
      for (int j = 0; j < (K); j++) {
        int mex = A[N] + j;
        int remSlots = N - mex;
        if (i > remSlots || remSlots < 0) continue;
        ans += choose(remSlots, i) * cur[i][j] * facs[i];
      }
    }
    cout << ans << nl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    initFacs();
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  