#include <bits/stdc++.h>
int quick_power(int a, int b, int Mod) {
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1ll * ans * a % Mod;
    }
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ans;
}
const int Maxn = 2000;
const int Maxk = 50;
const int Mod = 998244353;
int find_inv(int x) { return quick_power(x, Mod - 2, Mod); }
int f[2][Maxk * 2 + 5][Maxn + 5];
int n, k;
int a[Maxn + 5], b[Maxn + 5];
int fac[Maxn + 5], inv_f[Maxn + 5];
void init() {
  fac[0] = 1;
  for (int i = 1; i <= Maxn; i++) {
    fac[i] = 1ll * fac[i - 1] * i % Mod;
  }
  inv_f[Maxn] = find_inv(fac[Maxn]);
  for (int i = Maxn - 1; i >= 0; i--) {
    inv_f[i] = 1ll * inv_f[i + 1] * (i + 1) % Mod;
  }
}
int sum[Maxn * 2 + 5];
int main() {
  init();
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    a[i] = std::max(b[i] - k, 0);
    b[i] = std::min(b[i] + k, i);
  }
  int cur = 0, nxt;
  f[cur][0][0] = 1;
  for (int i = 1; i <= n; i++) {
    nxt = cur ^ 1;
    memset(f[nxt], 0, sizeof f[nxt]);
    memset(sum, 0, sizeof sum);
    for (int j = a[i], p = a[i - 1] - 1; j <= b[i]; j++) {
      while (p < b[i - 1] && p < j - 1) {
        p++;
        for (int k = 0; k < i; k++) {
          sum[p + k] =
              (sum[p + k] + 1ll * fac[k] * f[cur][p - a[i - 1]][k]) % Mod;
        }
      }
      for (int k = 0; k <= i; k++) {
        int id = j - a[i];
        f[nxt][id][k] = 1ll * sum[j + k - 1] * inv_f[k] % Mod;
        if (a[i - 1] <= j && j <= b[i - 1]) {
          f[nxt][id][k] = (f[nxt][id][k] + f[cur][j - a[i - 1]][k - 1]) % Mod;
          f[nxt][id][k] =
              (f[nxt][id][k] + 1ll * f[cur][j - a[i - 1]][k] * (j + k)) % Mod;
        }
      }
    }
    cur = nxt;
  }
  int ans = 0;
  for (int i = a[n]; i <= b[n]; i++) {
    for (int j = 0; i + j <= n; j++) {
      ans = (ans +
             1ll * f[cur][i - a[n]][j] * fac[n - i] % Mod * inv_f[n - i - j]) %
            Mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}

### 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>
int quick_power(int a, int b, int Mod) {
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1ll * ans * a % Mod;
    }
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ans;
}
const int Maxn = 2000;
const int Maxk = 50;
const int Mod = 998244353;
int find_inv(int x) { return quick_power(x, Mod - 2, Mod); }
int f[2][Maxk * 2 + 5][Maxn + 5];
int n, k;
int a[Maxn + 5], b[Maxn + 5];
int fac[Maxn + 5], inv_f[Maxn + 5];
void init() {
  fac[0] = 1;
  for (int i = 1; i <= Maxn; i++) {
    fac[i] = 1ll * fac[i - 1] * i % Mod;
  }
  inv_f[Maxn] = find_inv(fac[Maxn]);
  for (int i = Maxn - 1; i >= 0; i--) {
    inv_f[i] = 1ll * inv_f[i + 1] * (i + 1) % Mod;
  }
}
int sum[Maxn * 2 + 5];
int main() {
  init();
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    a[i] = std::max(b[i] - k, 0);
    b[i] = std::min(b[i] + k, i);
  }
  int cur = 0, nxt;
  f[cur][0][0] = 1;
  for (int i = 1; i <= n; i++) {
    nxt = cur ^ 1;
    memset(f[nxt], 0, sizeof f[nxt]);
    memset(sum, 0, sizeof sum);
    for (int j = a[i], p = a[i - 1] - 1; j <= b[i]; j++) {
      while (p < b[i - 1] && p < j - 1) {
        p++;
        for (int k = 0; k < i; k++) {
          sum[p + k] =
              (sum[p + k] + 1ll * fac[k] * f[cur][p - a[i - 1]][k]) % Mod;
        }
      }
      for (int k = 0; k <= i; k++) {
        int id = j - a[i];
        f[nxt][id][k] = 1ll * sum[j + k - 1] * inv_f[k] % Mod;
        if (a[i - 1] <= j && j <= b[i - 1]) {
          f[nxt][id][k] = (f[nxt][id][k] + f[cur][j - a[i - 1]][k - 1]) % Mod;
          f[nxt][id][k] =
              (f[nxt][id][k] + 1ll * f[cur][j - a[i - 1]][k] * (j + k)) % Mod;
        }
      }
    }
    cur = nxt;
  }
  int ans = 0;
  for (int i = a[n]; i <= b[n]; i++) {
    for (int j = 0; i + j <= n; j++) {
      ans = (ans +
             1ll * f[cur][i - a[n]][j] * fac[n - i] % Mod * inv_f[n - i - j]) %
            Mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```