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;
  const int N = 2005;
  int b[N], n, k;
  int f[2][105][N], g[2][105][N];
  long long jc[N * 2], inv[N * 2];
  int main() {
    scanf("%d %d", &n, &k);
    for (int i = (1), e_ = (n); i <= e_; i++) scanf("%d", &b[i]), b[i] -= k;
    k += k;
    jc[0] = inv[0] = jc[1] = inv[1] = 1;
    for (int i = (2), e_ = (n + n + 2); i <= e_; i++)
      jc[i] = jc[i - 1] * i % 998244353,
      inv[i] = inv[998244353 % i] * (998244353 - 998244353 / i) % 998244353;
    for (int i = (2), e_ = (n + n + 2); i <= e_; i++)
      inv[i] = inv[i - 1] * inv[i] % 998244353;
    long long ans = 0;
    f[0][0][0] = 1;
    for (int i = (0), e_ = (n); i <= e_; i++) {
      int u = (i & 1), v = u ^ 1;
      for (int j = (0), e_ = (k + 1); j <= e_; j++)
        memset(f[v][j], 0, sizeof(f[v][j][0]) * (i + 2)),
            memset(g[v][j], 0, sizeof(g[v][j][0]) * (i + 2));
      for (int j = (0), e_ = (k); j <= e_; j++)
        for (int t = (0), e_ = (i); t <= e_; t++)
          if (g[u][j][t]) {
            ((f[u][j][t] += g[u][j][t]) &&
             (f[u][j][t] -= (f[u][j][t] >= 998244353) ? 998244353 : 0));
            if (t && j < k)
              (g[u][j + 1][t - 1] =
                   ((1ll * g[u][j][t] * t) + (g[u][j + 1][t - 1])) % 998244353);
          }
      for (int j = (0), e_ = (k); j <= e_; j++)
        for (int t = (0), e_ = (i); t <= e_; t++)
          if (f[u][j][t]) {
            if (i == n) {
              int d = n - b[i] - j;
              if (d >= t)
                (ans = ((1ll * f[u][j][t] * jc[d] % 998244353 * inv[d - t]) +
                        (ans)) %
                       998244353);
              continue;
            }
            int x = j - (b[i + 1] - b[i]);
            if (x > k) continue;
            if (x >= 0)
              (f[v][x][t] = ((1ll * f[u][j][t] * (b[i] + j + t)) + (f[v][x][t])) %
                            998244353),
                  ((f[v][x][t + 1] += f[u][j][t]) &&
                   (f[v][x][t + 1] -=
                    (f[v][x][t + 1] >= 998244353) ? 998244353 : 0));
            if (x < -1) {
              int d = b[i + 1] - (b[i] + j) - 1;
              if (t >= d)
                (g[v][0][t - d] =
                     ((1ll * f[u][j][t] * jc[t] % 998244353 * inv[t - d]) +
                      (g[v][0][t - d])) %
                     998244353);
            } else
              ((g[v][x + 1][t] += f[u][j][t]) &&
               (g[v][x + 1][t] -= (g[v][x + 1][t] >= 998244353) ? 998244353 : 0));
          }
    }
    printf("%lld", ans);
    return 0;
  }
  