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, K = 55, mod = 998244353;
  int n, k, b[N], f[2 * K][N], g[2 * K][N], C[N][N];
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
    f[k][0] = 1;
    for (int i = 1; i <= n; i++) {
      memset(g, 0, sizeof(g));
      for (int j = max(0, k - b[i - 1]); j <= 2 * k; j++) {
        int j2 = j + b[i - 1] - b[i] + 1, x = j + b[i - 1] - k;
        if (x < 0 || x > n || j2 > 2 * k) continue;
        for (int d = max(0, -j2); d < i; d++) {
          if (j2 < 0)
            (g[0][d + j2] += f[j][d]) %= mod;
          else
            (g[j2][d] += f[j][d]) %= mod;
        }
      }
      for (int j = 0; j < 2 * k; j++)
        for (int d = 1; d <= i; d++) (g[j + 1][d - 1] += g[j][d]) %= mod;
      for (int j = max(0, k - b[i - 1]); j <= 2 * k; j++) {
        int j2 = j + b[i - 1] - b[i], x = j + b[i - 1] - k;
        if (x < 0 || x > n || j2 < 0 || j2 > 2 * k) continue;
        for (int d = 0; d < i; d++) {
          (g[j2][d] += 1ll * f[j][d] * (d + x) % mod) %= mod;
          (g[j2][d + 1] += 1ll * f[j][d] * (d + 1) % mod) %= mod;
        }
      }
      swap(f, g);
    }
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      C[i][0] = C[i][i] = 1;
      for (int j = 1; j < n; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
    int ans = 0;
    for (int j = 0; j <= 2 * k; j++) {
      int x = b[n] + j - k;
      if (x < 0 || x > n) continue;
      for (int d = 0; d <= n - x; d++)
        (ans += 1ll * f[j][d] * C[n - x][d] % mod) %= mod;
    }
    printf("%d\n", ans);
    return 0;
  }
  