#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const int M = 998244353;
int n, K, b[2003];
int f[105][2003], nf[105][2003], s[105][2003];
int L[2003], R[2003], fc[2003], inv[2003];
int myp(int x, int t) {
  int ans = 1;
  for (; t; t >>= 1, x = (long long)x * x % M)
    if (t & 1) ans = (long long)ans * x % M;
  return ans;
}
int main() {
  fc[0] = inv[0] = 1;
  for (int i = 1; i < 2003; i++)
    fc[i] = (long long)fc[i - 1] * i % M, inv[i] = myp(fc[i], M - 2);
  scanf("%d%d", &n, &K);
  b[0] = -K;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 0; i <= n; i++) L[i] = max(b[i] - K, 0), R[i] = b[i] + K;
  f[0][0] = 1;
  for (int i = 0; i < n; i++) {
    memset(nf, 0, sizeof(nf));
    memset(s, 0, sizeof(s));
    for (int j = L[i]; j <= R[i]; j++) {
      for (int l = 0; l <= i; l++)
        if (f[j - L[i]][l]) {
          if (abs(j - b[i]) <= K) {
            (nf[j - L[i + 1]][l + 1] += f[j - L[i]][l]) %= M;
            (nf[j - L[i + 1]][l] += (long long)f[j - L[i]][l] * (j + l) % M) %=
                M;
          }
          if (j + 1 <= R[i + 1])
            (s[max(j + 1, L[i + 1]) - L[i + 1]][j + l] +=
             (long long)fc[l] * f[j - L[i]][l] % M) %= M;
        }
    }
    for (int T = L[i]; T <= i; T++)
      for (int j = L[i + 1]; j <= R[i + 1]; j++) {
        if (j > L[i + 1]) (s[j - L[i + 1]][T] += s[j - 1 - L[i + 1]][T]) %= M;
        (nf[j - L[i + 1]][T + 1 - j] +=
         (long long)s[j - L[i + 1]][T] * inv[T + 1 - j] % M) %= M;
      }
    memcpy(f, nf, sizeof(nf));
  }
  int ans = 0;
  for (int i = L[n]; i <= R[n]; i++)
    for (int j = 0; j <= n; j++)
      if (j <= n - i)
        (ans +=
         (long long)f[i - L[n]][j] * fc[n - i] % M * inv[n - i - j] % M) %= M;
  printf("%d", ans);
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const int M = 998244353;
int n, K, b[2003];
int f[105][2003], nf[105][2003], s[105][2003];
int L[2003], R[2003], fc[2003], inv[2003];
int myp(int x, int t) {
  int ans = 1;
  for (; t; t >>= 1, x = (long long)x * x % M)
    if (t & 1) ans = (long long)ans * x % M;
  return ans;
}
int main() {
  fc[0] = inv[0] = 1;
  for (int i = 1; i < 2003; i++)
    fc[i] = (long long)fc[i - 1] * i % M, inv[i] = myp(fc[i], M - 2);
  scanf("%d%d", &n, &K);
  b[0] = -K;
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 0; i <= n; i++) L[i] = max(b[i] - K, 0), R[i] = b[i] + K;
  f[0][0] = 1;
  for (int i = 0; i < n; i++) {
    memset(nf, 0, sizeof(nf));
    memset(s, 0, sizeof(s));
    for (int j = L[i]; j <= R[i]; j++) {
      for (int l = 0; l <= i; l++)
        if (f[j - L[i]][l]) {
          if (abs(j - b[i]) <= K) {
            (nf[j - L[i + 1]][l + 1] += f[j - L[i]][l]) %= M;
            (nf[j - L[i + 1]][l] += (long long)f[j - L[i]][l] * (j + l) % M) %=
                M;
          }
          if (j + 1 <= R[i + 1])
            (s[max(j + 1, L[i + 1]) - L[i + 1]][j + l] +=
             (long long)fc[l] * f[j - L[i]][l] % M) %= M;
        }
    }
    for (int T = L[i]; T <= i; T++)
      for (int j = L[i + 1]; j <= R[i + 1]; j++) {
        if (j > L[i + 1]) (s[j - L[i + 1]][T] += s[j - 1 - L[i + 1]][T]) %= M;
        (nf[j - L[i + 1]][T + 1 - j] +=
         (long long)s[j - L[i + 1]][T] * inv[T + 1 - j] % M) %= M;
      }
    memcpy(f, nf, sizeof(nf));
  }
  int ans = 0;
  for (int i = L[n]; i <= R[n]; i++)
    for (int j = 0; j <= n; j++)
      if (j <= n - i)
        (ans +=
         (long long)f[i - L[n]][j] * fc[n - i] % M * inv[n - i - j] % M) %= M;
  printf("%d", ans);
}

```