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 mod = 998244353;
  inline int addmod(int x) { return x >= mod ? x - mod : x; }
  int n, m, b[100005], f[2][2005][2005], c[2005][2005], pn = 2e3;
  inline void upd(int &x, int y) { x = addmod(x + y); }
  int main() {
    for (int i = 0; i <= pn; i++) c[i][0] = 1;
    for (int i = 1; i <= pn; i++)
      for (int j = 1; j <= pn; j++)
        c[i][j] = addmod(c[i - 1][j - 1] + c[i - 1][j]);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
    f[0][0][0] = 1;
    int L = 0;
    for (int i = 0; i < n; i++) {
      int nw = i % 2;
      for (int j = 0; j <= n; j++)
        if (min(min(L - 1, b[i] - m), b[i + 1] - m) <= j && j <= b[i + 1] + m)
          for (int k = 0; k <= i + 1; k++) f[nw ^ 1][j][k] = 0;
      for (int j = 0; j <= n; j++)
        if (b[i] - m <= j && j <= b[i] + m)
          for (int k = 0; k <= i + 1; k++) upd(f[nw ^ 1][j + 1][k], f[nw][j][k]);
      for (int j = 0; j <= n; j++)
        if (L <= j + 1 && j + 1 <= b[i + 1] + m)
          for (int k = 1; k <= i + 1; k++)
            upd(f[nw ^ 1][j + 1][k - 1], f[nw ^ 1][j][k]);
      for (int j = 0; j <= n; j++)
        if (b[i] - m <= j && j <= b[i] + m)
          for (int k = 0; k <= i + 1; k++) {
            upd(f[nw ^ 1][j][k], 1ll * (k + j) * f[nw][j][k] % mod);
            upd(f[nw ^ 1][j][k + 1], 1ll * (k + 1) * f[nw][j][k] % mod);
          }
      L = max(L, b[i] - m);
    }
    int ans = 0;
    for (int i = 0; i <= n; i++)
      if (b[n] - m <= i && i <= b[n] + m)
        for (int j = 0; j <= n; j++)
          upd(ans, 1ll * f[n % 2][i][j] * c[n - i][j] % mod);
    printf("%d\n", ans);
    return 0;
  }
  