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>
  int quick_power(int a, int b, int Mod) {
    int ans = 1;
    while (b) {
      if (b & 1) {
        ans = 1ll * ans * a % Mod;
      }
      b >>= 1;
      a = 1ll * a * a % Mod;
    }
    return ans;
  }
  const int Maxn = 2000;
  const int Maxk = 50;
  const int Mod = 998244353;
  int find_inv(int x) { return quick_power(x, Mod - 2, Mod); }
  int f[2][Maxk * 2 + 5][Maxn + 5];
  int n, k;
  int a[Maxn + 5], b[Maxn + 5];
  int fac[Maxn + 5], inv_f[Maxn + 5];
  void init() {
    fac[0] = 1;
    for (int i = 1; i <= Maxn; i++) {
      fac[i] = 1ll * fac[i - 1] * i % Mod;
    }
    inv_f[Maxn] = find_inv(fac[Maxn]);
    for (int i = Maxn - 1; i >= 0; i--) {
      inv_f[i] = 1ll * inv_f[i + 1] * (i + 1) % Mod;
    }
  }
  int sum[Maxn * 2 + 5];
  int main() {
    init();
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      a[i] = std::max(b[i] - k, 0);
      b[i] = std::min(b[i] + k, i);
    }
    int cur = 0, nxt;
    f[cur][0][0] = 1;
    for (int i = 1; i <= n; i++) {
      nxt = cur ^ 1;
      memset(f[nxt], 0, sizeof f[nxt]);
      memset(sum, 0, sizeof sum);
      for (int j = a[i], p = a[i - 1] - 1; j <= b[i]; j++) {
        while (p < b[i - 1] && p < j - 1) {
          p++;
          for (int k = 0; k < i; k++) {
            sum[p + k] =
                (sum[p + k] + 1ll * fac[k] * f[cur][p - a[i - 1]][k]) % Mod;
          }
        }
        for (int k = 0; k <= i; k++) {
          int id = j - a[i];
          f[nxt][id][k] = 1ll * sum[j + k - 1] * inv_f[k] % Mod;
          if (a[i - 1] <= j && j <= b[i - 1]) {
            f[nxt][id][k] = (f[nxt][id][k] + f[cur][j - a[i - 1]][k - 1]) % Mod;
            f[nxt][id][k] =
                (f[nxt][id][k] + 1ll * f[cur][j - a[i - 1]][k] * (j + k)) % Mod;
          }
        }
      }
      cur = nxt;
    }
    int ans = 0;
    for (int i = a[n]; i <= b[n]; i++) {
      for (int j = 0; i + j <= n; j++) {
        ans = (ans +
               1ll * f[cur][i - a[n]][j] * fac[n - i] % Mod * inv_f[n - i - j]) %
              Mod;
      }
    }
    printf("%d\n", ans);
    return 0;
  }
  