#include <bits/stdc++.h>
using namespace std;
const int N = 2005, M = 55, mod = 998244353;
int n, m, b[N], st[N], ed[N], f[M << 1][N], g[M << 1][N], sum[N << 1], fac[N],
    inv[N];
int fpow_(int a, int b, int res = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) {
      res = 1ll * res * a % mod;
    }
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    st[i] = max(b[i] - m, 0), ed[i] = min(b[i] + m, i);
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * i * fac[i - 1] % mod;
  }
  inv[n] = fpow_(fac[n], mod - 2);
  for (int i = n; i; i--) {
    inv[i - 1] = 1ll * i * inv[i] % mod;
  }
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    memcpy(g, f, sizeof(f));
    memset(f, 0, sizeof(f));
    memset(sum, 0, sizeof(sum));
    for (int j = st[i], p = st[i - 1] - 1; j <= ed[i]; j++) {
      while (p < ed[i - 1] && p < j - 1) {
        p++;
        for (int k = 0; k < i; k++) {
          sum[p + k] = (sum[p + k] + 1ll * fac[k] * g[p - st[i - 1]][k]) % mod;
        }
      }
      for (int k = 0; k <= i; k++) {
        int x = j - st[i];
        f[x][k] = 1ll * sum[j + k - 1] * inv[k] % mod;
        if (st[i - 1] <= j && j <= ed[i - 1]) {
          f[x][k] = (f[x][k] + g[j - st[i - 1]][k - 1]) % mod;
          f[x][k] = (f[x][k] + 1ll * g[j - st[i - 1]][k] * (j + k)) % mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = st[n]; i <= ed[n]; i++) {
    for (int j = 0; i + j <= n; j++) {
      ans = (ans + 1ll * f[i - st[n]][j] * fac[n - i] % mod * inv[n - i - j]) %
            mod;
    }
  }
  printf("%d\n", (ans % mod + mod) % mod);
}

### Prompt

Generate a Cpp solution 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 N = 2005, M = 55, mod = 998244353;
int n, m, b[N], st[N], ed[N], f[M << 1][N], g[M << 1][N], sum[N << 1], fac[N],
    inv[N];
int fpow_(int a, int b, int res = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) {
      res = 1ll * res * a % mod;
    }
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    st[i] = max(b[i] - m, 0), ed[i] = min(b[i] + m, i);
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = 1ll * i * fac[i - 1] % mod;
  }
  inv[n] = fpow_(fac[n], mod - 2);
  for (int i = n; i; i--) {
    inv[i - 1] = 1ll * i * inv[i] % mod;
  }
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    memcpy(g, f, sizeof(f));
    memset(f, 0, sizeof(f));
    memset(sum, 0, sizeof(sum));
    for (int j = st[i], p = st[i - 1] - 1; j <= ed[i]; j++) {
      while (p < ed[i - 1] && p < j - 1) {
        p++;
        for (int k = 0; k < i; k++) {
          sum[p + k] = (sum[p + k] + 1ll * fac[k] * g[p - st[i - 1]][k]) % mod;
        }
      }
      for (int k = 0; k <= i; k++) {
        int x = j - st[i];
        f[x][k] = 1ll * sum[j + k - 1] * inv[k] % mod;
        if (st[i - 1] <= j && j <= ed[i - 1]) {
          f[x][k] = (f[x][k] + g[j - st[i - 1]][k - 1]) % mod;
          f[x][k] = (f[x][k] + 1ll * g[j - st[i - 1]][k] * (j + k)) % mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = st[n]; i <= ed[n]; i++) {
    for (int j = 0; i + j <= n; j++) {
      ans = (ans + 1ll * f[i - st[n]][j] * fac[n - i] % mod * inv[n - i - j]) %
            mod;
    }
  }
  printf("%d\n", (ans % mod + mod) % mod);
}

```