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 <typename T>
  void chmin(T& x, const T& y) {
    if (x > y) x = y;
  }
  template <typename T>
  void chmax(T& x, const T& y) {
    if (x < y) x = y;
  }
  char readc() {
    char c;
    while (isspace((c = getchar())))
      ;
    return c;
  }
  int read() {
    char c;
    while ((c = getchar()) < '-')
      ;
    if (c == '-') {
      int x = (c = getchar()) - '0';
      while ((c = getchar()) >= '0') x = x * 10 + c - '0';
      return -x;
    }
    int x = c - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return x;
  }
  const int N = 2000 + 50, K = 55, D = 998244353;
  int l[N], r[N];
  int64_t dp[2][K * 2][N], C[N][N], fact[N], P[N][N];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    for (int i = 0; i < N; ++i) {
      C[i][0] = 1;
      for (int j = 1; j <= i; ++j) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % D;
    }
    fact[0] = 1;
    for (int i = 1; i < N; ++i) fact[i] = fact[i - 1] * i % D;
    for (int i = 0; i < N; ++i)
      for (int j = 0; j <= i; ++j) P[i][j] = C[i][j] * fact[j] % D;
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
      cin >> r[i];
      l[i] = max(r[i] - k, l[i - 1]);
      r[i] = min(r[i] + k + 1, n + 1);
      if (r[i] <= l[i]) {
        puts("0");
        exit(0);
      }
    }
    bool d = 0;
    dp[d][0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      d ^= 1;
      memset(dp[d], 0, sizeof(dp[d]));
      for (int j = l[i - 1]; j <= l[i - 1] + 2 * k + 1; ++j)
        for (int k = 0; k < i; ++k) {
          int64_t f0 = dp[d ^ 1][j - l[i - 1]][k];
          if (!f0) continue;
          auto trans = [&](int nj) {
            if (nj <= j) {
              (dp[d][j - l[i]][k] += f0 * (j + k)) %= D;
              (dp[d][j - l[i]][k + 1] += f0) %= D;
            } else {
              if (k >= nj - j)
                (dp[d][nj - l[i]][k - (nj - j)] +=
                 f0 * (j + k) % D * P[k][nj - j]) %= D;
              if (k + 1 >= nj - j)
                (dp[d][nj - l[i]][k + 1 - (nj - j)] += f0 * P[k + 1][nj - j]) %=
                    D;
            }
          };
          trans(l[i]);
          f0 = -f0;
          trans(r[i]);
        }
    }
    int64_t ans = 0;
    for (int j = l[n]; j <= l[n] + 2 * k + 1; ++j)
      for (int k = 0; k <= n + 1 - j; ++k) {
        (ans += dp[d][j - l[n]][k] * P[n + 1 - j][k]) %= D;
      }
    cout << (ans % D + D) % D << endl;
  }
  