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, M = 55, mod = 998244353;
  int n, m, b[N], st[N], ed[N], f[M << 1][N], g[M << 1][N], sum[N << 1], fac[N],
      inv[N];
  int fpow_(int a, int b, int res = 1) {
    for (; b; b >>= 1, a = 1ll * a * a % mod) {
      if (b & 1) {
        res = 1ll * res * a % mod;
      }
    }
    return res;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      st[i] = max(b[i] - m, 0), ed[i] = min(b[i] + m, i);
    }
    fac[0] = 1;
    for (int i = 1; i <= n; i++) {
      fac[i] = 1ll * i * fac[i - 1] % mod;
    }
    inv[n] = fpow_(fac[n], mod - 2);
    for (int i = n; i; i--) {
      inv[i - 1] = 1ll * i * inv[i] % mod;
    }
    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      memcpy(g, f, sizeof(f));
      memset(f, 0, sizeof(f));
      memset(sum, 0, sizeof(sum));
      for (int j = st[i], p = st[i - 1] - 1; j <= ed[i]; j++) {
        while (p < ed[i - 1] && p < j - 1) {
          p++;
          for (int k = 0; k < i; k++) {
            sum[p + k] = (sum[p + k] + 1ll * fac[k] * g[p - st[i - 1]][k]) % mod;
          }
        }
        for (int k = 0; k <= i; k++) {
          int x = j - st[i];
          f[x][k] = 1ll * sum[j + k - 1] * inv[k] % mod;
          if (st[i - 1] <= j && j <= ed[i - 1]) {
            f[x][k] = (f[x][k] + g[j - st[i - 1]][k - 1]) % mod;
            f[x][k] = (f[x][k] + 1ll * g[j - st[i - 1]][k] * (j + k)) % mod;
          }
        }
      }
    }
    int ans = 0;
    for (int i = st[n]; i <= ed[n]; i++) {
      for (int j = 0; i + j <= n; j++) {
        ans = (ans + 1ll * f[i - st[n]][j] * fac[n - i] % mod * inv[n - i - j]) %
              mod;
      }
    }
    printf("%d\n", (ans % mod + mod) % mod);
  }
  