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 P = 998244353;
  struct Z {
    int x;
    Z(int _x = 0) : x(_x) {}
    inline Z operator-() const {
      if (!x) return 0;
      return Z(P - x);
    }
    inline Z &operator+=(const Z &rhs) {
      x += rhs.x;
      if (x >= P) x -= P;
      return *this;
    }
    inline Z &operator-=(const Z &rhs) {
      x -= rhs.x;
      if (x < 0) x += P;
      return *this;
    }
    inline Z &operator*=(const Z &rhs) {
      x = 1ull * x * rhs.x % P;
      return *this;
    }
    inline Z operator+(const Z &rhs) const {
      Z res = *this;
      return res += rhs;
    }
    inline Z operator-(const Z &rhs) const {
      Z res = *this;
      return res -= rhs;
    }
    inline Z operator*(const Z &rhs) const {
      Z res = *this;
      return res *= rhs;
    }
  };
  int n, k, b[2005];
  Z f[105][2005], g[105][2005], sum[2005];
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
    b[0] = -k;
    for (int j = 0; j <= 2 * k; ++j) {
      int val = b[n] + j - k;
      if (val >= 0 && val <= n) f[j][1] = 1;
    }
    for (int i = n; i; --i) {
      memset(g, 0, sizeof g);
      for (int j = 0; j <= 2 * k; ++j) {
        int pos = b[i] + j - k - b[i - 1] + k;
        if (pos < 0 || pos > 2 * k) continue;
        for (int c = 1; c <= n + 1; ++c) g[pos][c] += f[j][c] * (n + 1 - c);
      }
      memset(sum, 0, sizeof sum);
      for (int v = b[i] + k - 1; v >= b[i - 1] - k; --v) {
        for (int c = n + 1; c; --c) sum[c] -= sum[c - 1];
        int pos = v + 1 - b[i] + k;
        if (pos >= 0) {
          for (int c = 1; c <= n + 1; ++c) sum[c] += f[pos][c];
        }
        pos = v - b[i - 1] + k;
        if (pos <= 2 * k) {
          for (int c = 1; c <= n; ++c) g[pos][c + 1] += sum[c];
        }
      }
      memcpy(f, g, sizeof f);
    }
    Z ans = 0;
    for (int c = 1; c <= n + 1; ++c) ans += f[2 * k][c];
    printf("%d\n", ans.x);
    return 0;
  }
  