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>
  const int N = 2005, K = 105, MOD = 998244353;
  inline int mval(int x) { return x >= MOD ? x - MOD : x; }
  inline int fix(int x) { return mval(x + MOD); }
  inline void inc(int &x, int a) { x = mval(x + a); }
  inline void dec(int &x, int a) { x = fix(x - a); }
  inline int qpow(int x, int p) {
    int ret = 1;
    while (p) {
      if (p & 1) ret = 1ll * ret * x % MOD;
      p >>= 1, x = 1ll * x * x % MOD;
    }
    return ret;
  }
  int n, k, f[K][N], g[K][N], fac[N], inv[N], b[N], l[N], r[N];
  inline int C(int n, int m) {
    return 1ll * fac[n] * inv[m] % MOD * inv[n - m] % MOD;
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", b + i);
      l[i] = b[i] - k;
      r[i] = b[i] + k;
      l[i] = std::max(l[i], l[i - 1]);
    }
    r[n + 1] = n + 1;
    for (int i = n; i; --i) r[i] = std::min(r[i], r[i + 1]);
    for (int i = 1; i <= n; ++i)
      if (l[i] > r[i]) return puts("0"), 0;
    inv[0] = inv[1] = fac[0] = 1;
    for (int i = 2; i <= n; ++i)
      inv[i] = MOD - 1ll * (MOD / i) * inv[MOD % i] % MOD;
    for (int i = 1; i <= n; ++i)
      inv[i] = 1ll * inv[i] * inv[i - 1] % MOD,
      fac[i] = 1ll * i * fac[i - 1] % MOD;
    for (int i = 0; i <= r[n] - l[n]; ++i)
      for (int j = l[n] + i; j <= n; ++j)
        f[i][j] = C(n - (l[n] + i), j - l[n] - i);
    if (r[n] > n) f[r[n] - l[n]][n] = 1;
    for (int i = n; i; --i) {
      memset(g, 0, sizeof(g));
      for (int j = 0; j <= r[i] - l[i]; ++j) {
        int rv = j + l[i];
        for (int s = std::max(1, l[i] + j); s <= i; ++s)
          if (f[j][s]) {
            int val = f[j][s];
            inc(g[std::min(r[i - 1], rv - 1) - l[i - 1]][s - 1], val);
          }
      }
      for (int j = r[i - 1] - l[i - 1]; ~j; --j)
        for (int s = l[i - 1] + j; s <= i; ++s) inc(g[j][s], g[j + 1][s]);
      for (int j = 0; j <= r[i] - l[i]; ++j) {
        int rv = j + l[i];
        for (int s = l[i] + j; s <= i; ++s)
          if (f[j][s]) {
            int val = f[j][s], t = rv - l[i - 1];
            if (s) inc(g[t][s - 1], 1ll * (s - rv) * val % MOD);
            inc(g[t][s], 1ll * s * val % MOD);
          }
      }
      memcpy(f, g, sizeof(f));
    }
    printf("%d\n", f[0][0]);
    return 0;
  }
  