#include <bits/stdc++.h>
using namespace std;
const int maxK = 111;
const int maxN = 2222;
const int mod = 998244353;
int fac[maxN], ifac[maxN], inv[maxN];
int L[maxN], R[maxN], f[2][maxN][maxK], g[2][maxN][maxK];
void init() {
  inv[1] = 1;
  fac[0] = ifac[0] = 1;
  fac[1] = ifac[1] = 1;
  for (int i = 2; i < maxN; i++) {
    fac[i] = i * 1LL * fac[i - 1] % mod;
    inv[i] = (mod - mod / i) * 1LL * inv[mod % i] % mod;
    ifac[i] = ifac[i - 1] * 1LL * inv[i] % mod;
  }
}
inline int npr(int n, int r) {
  if (r < 0 || r > n) return 0;
  return fac[n] * 1LL * ifac[n - r] % mod;
}
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  init();
  int N, K;
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    int x;
    cin >> x;
    L[i] = x - K;
    R[i] = x + K;
    if (L[i] < 0) L[i] = 0;
  }
  L[0] = R[0] = 0;
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  int flg = 0;
  f[flg][0][0] = 1;
  g[flg][0][0] = 1;
  for (int i = 1; i <= N; i++) {
    int nlg = flg ^ 1;
    memset(f[nlg], 0, sizeof(f[nlg]));
    memset(g[nlg], 0, sizeof(g[nlg]));
    for (int d = 0; d <= i; d++) {
      for (int y = L[i]; y <= R[i]; y++) {
        if (L[i - 1] <= y && y <= R[i - 1]) {
          add(f[nlg][d][y - L[i]],
              1LL * (d + y) * f[flg][d][y - L[i - 1]] % mod);
          if (d > 0) {
            add(f[nlg][d][y - L[i]], f[flg][d - 1][y - L[i - 1]]);
          }
        }
        if (y > L[i - 1]) {
          int ny = min(y - 1, R[i - 1]);
          int nd = d + y - 1 - ny;
          add(f[nlg][d][y - L[i]],
              1LL * ifac[d] * g[flg][nd][ny - L[i - 1]] % mod);
        }
      }
    }
    for (int d = i; d >= 0; d--) {
      for (int y = L[i]; y <= R[i]; y++) {
        g[nlg][d][y - L[i]] = fac[d] * 1LL * f[nlg][d][y - L[i]] % mod;
        if (y > L[i]) {
          add(g[nlg][d][y - L[i]], g[nlg][d + 1][y - L[i] - 1]);
        }
      }
    }
    flg = nlg;
  }
  int answer = 0;
  for (int d = 0; d <= N; d++) {
    for (int x = L[N]; x <= R[N]; x++) {
      add(answer, npr(N - x, d) * 1LL * f[flg][d][x - L[N]] % mod);
    }
  }
  cout << answer << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxK = 111;
const int maxN = 2222;
const int mod = 998244353;
int fac[maxN], ifac[maxN], inv[maxN];
int L[maxN], R[maxN], f[2][maxN][maxK], g[2][maxN][maxK];
void init() {
  inv[1] = 1;
  fac[0] = ifac[0] = 1;
  fac[1] = ifac[1] = 1;
  for (int i = 2; i < maxN; i++) {
    fac[i] = i * 1LL * fac[i - 1] % mod;
    inv[i] = (mod - mod / i) * 1LL * inv[mod % i] % mod;
    ifac[i] = ifac[i - 1] * 1LL * inv[i] % mod;
  }
}
inline int npr(int n, int r) {
  if (r < 0 || r > n) return 0;
  return fac[n] * 1LL * ifac[n - r] % mod;
}
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  init();
  int N, K;
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    int x;
    cin >> x;
    L[i] = x - K;
    R[i] = x + K;
    if (L[i] < 0) L[i] = 0;
  }
  L[0] = R[0] = 0;
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  int flg = 0;
  f[flg][0][0] = 1;
  g[flg][0][0] = 1;
  for (int i = 1; i <= N; i++) {
    int nlg = flg ^ 1;
    memset(f[nlg], 0, sizeof(f[nlg]));
    memset(g[nlg], 0, sizeof(g[nlg]));
    for (int d = 0; d <= i; d++) {
      for (int y = L[i]; y <= R[i]; y++) {
        if (L[i - 1] <= y && y <= R[i - 1]) {
          add(f[nlg][d][y - L[i]],
              1LL * (d + y) * f[flg][d][y - L[i - 1]] % mod);
          if (d > 0) {
            add(f[nlg][d][y - L[i]], f[flg][d - 1][y - L[i - 1]]);
          }
        }
        if (y > L[i - 1]) {
          int ny = min(y - 1, R[i - 1]);
          int nd = d + y - 1 - ny;
          add(f[nlg][d][y - L[i]],
              1LL * ifac[d] * g[flg][nd][ny - L[i - 1]] % mod);
        }
      }
    }
    for (int d = i; d >= 0; d--) {
      for (int y = L[i]; y <= R[i]; y++) {
        g[nlg][d][y - L[i]] = fac[d] * 1LL * f[nlg][d][y - L[i]] % mod;
        if (y > L[i]) {
          add(g[nlg][d][y - L[i]], g[nlg][d + 1][y - L[i] - 1]);
        }
      }
    }
    flg = nlg;
  }
  int answer = 0;
  for (int d = 0; d <= N; d++) {
    for (int x = L[N]; x <= R[N]; x++) {
      add(answer, npr(N - x, d) * 1LL * f[flg][d][x - L[N]] % mod);
    }
  }
  cout << answer << '\n';
  return 0;
}

```