#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
inline int addmod(int x) { return x >= mod ? x - mod : x; }
int n, m, b[100005], f[2][2005][2005], c[2005][2005], pn = 2e3;
inline void upd(int &x, int y) { x = addmod(x + y); }
int main() {
  for (int i = 0; i <= pn; i++) c[i][0] = 1;
  for (int i = 1; i <= pn; i++)
    for (int j = 1; j <= pn; j++)
      c[i][j] = addmod(c[i - 1][j - 1] + c[i - 1][j]);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  f[0][0][0] = 1;
  int L = 0;
  for (int i = 0; i < n; i++) {
    int nw = i % 2;
    for (int j = 0; j <= n; j++)
      if (min(min(L - 1, b[i] - m), b[i + 1] - m) <= j && j <= b[i + 1] + m)
        for (int k = 0; k <= i + 1; k++) f[nw ^ 1][j][k] = 0;
    for (int j = 0; j <= n; j++)
      if (b[i] - m <= j && j <= b[i] + m)
        for (int k = 0; k <= i + 1; k++) upd(f[nw ^ 1][j + 1][k], f[nw][j][k]);
    for (int j = 0; j <= n; j++)
      if (L <= j + 1 && j + 1 <= b[i + 1] + m)
        for (int k = 1; k <= i + 1; k++)
          upd(f[nw ^ 1][j + 1][k - 1], f[nw ^ 1][j][k]);
    for (int j = 0; j <= n; j++)
      if (b[i] - m <= j && j <= b[i] + m)
        for (int k = 0; k <= i + 1; k++) {
          upd(f[nw ^ 1][j][k], 1ll * (k + j) * f[nw][j][k] % mod);
          upd(f[nw ^ 1][j][k + 1], 1ll * (k + 1) * f[nw][j][k] % mod);
        }
    L = max(L, b[i] - m);
  }
  int ans = 0;
  for (int i = 0; i <= n; i++)
    if (b[n] - m <= i && i <= b[n] + m)
      for (int j = 0; j <= n; j++)
        upd(ans, 1ll * f[n % 2][i][j] * c[n - i][j] % mod);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to 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 mod = 998244353;
inline int addmod(int x) { return x >= mod ? x - mod : x; }
int n, m, b[100005], f[2][2005][2005], c[2005][2005], pn = 2e3;
inline void upd(int &x, int y) { x = addmod(x + y); }
int main() {
  for (int i = 0; i <= pn; i++) c[i][0] = 1;
  for (int i = 1; i <= pn; i++)
    for (int j = 1; j <= pn; j++)
      c[i][j] = addmod(c[i - 1][j - 1] + c[i - 1][j]);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  f[0][0][0] = 1;
  int L = 0;
  for (int i = 0; i < n; i++) {
    int nw = i % 2;
    for (int j = 0; j <= n; j++)
      if (min(min(L - 1, b[i] - m), b[i + 1] - m) <= j && j <= b[i + 1] + m)
        for (int k = 0; k <= i + 1; k++) f[nw ^ 1][j][k] = 0;
    for (int j = 0; j <= n; j++)
      if (b[i] - m <= j && j <= b[i] + m)
        for (int k = 0; k <= i + 1; k++) upd(f[nw ^ 1][j + 1][k], f[nw][j][k]);
    for (int j = 0; j <= n; j++)
      if (L <= j + 1 && j + 1 <= b[i + 1] + m)
        for (int k = 1; k <= i + 1; k++)
          upd(f[nw ^ 1][j + 1][k - 1], f[nw ^ 1][j][k]);
    for (int j = 0; j <= n; j++)
      if (b[i] - m <= j && j <= b[i] + m)
        for (int k = 0; k <= i + 1; k++) {
          upd(f[nw ^ 1][j][k], 1ll * (k + j) * f[nw][j][k] % mod);
          upd(f[nw ^ 1][j][k + 1], 1ll * (k + 1) * f[nw][j][k] % mod);
        }
    L = max(L, b[i] - m);
  }
  int ans = 0;
  for (int i = 0; i <= n; i++)
    if (b[n] - m <= i && i <= b[n] + m)
      for (int j = 0; j <= n; j++)
        upd(ans, 1ll * f[n % 2][i][j] * c[n - i][j] % mod);
  printf("%d\n", ans);
  return 0;
}

```