#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
template <typename T>
void write(T t) {
  if (t < 0) {
    putchar('-');
    write(-t);
    return;
  }
  if (t > 9) write(t / 10);
  putchar('0' + t % 10);
}
template <typename T>
void writeln(T t) {
  write(t);
  puts("");
}
const int mod = 998244353;
const int maxn = 2010;
int n, k, jc[maxn], ivjc[maxn], iv[maxn];
int dp[2][maxn][maxn], L[maxn], R[maxn], ans, tmp[maxn][maxn * 2];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int C(int x, int y) {
  if (x < y || y < 0) return 0;
  return (long long)jc[x] * ivjc[y] % mod * ivjc[x - y] % mod;
}
int main() {
  read(n), read(k);
  iv[0] = jc[0] = ivjc[0] = iv[1] = jc[1] = ivjc[1] = 1;
  for (int i = 2; i < maxn; i++) {
    jc[i] = (long long)jc[i - 1] * i % mod;
    iv[i] = (long long)iv[mod % i] * (mod - mod / i) % mod;
    ivjc[i] = (long long)ivjc[i - 1] * iv[i] % mod;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    L[i] = max(0, x - k), R[i] = min(n + 1, x + k);
  }
  dp[0][0][0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    for (int mex = L[i]; mex <= R[i]; ++mex)
      for (int cnt = 0; cnt <= i; ++cnt) dp[i & 1][mex][cnt] = 0;
    for (int mex = L[i - 1]; mex <= R[i - 1]; ++mex) {
      if (i <= n) {
        memset(tmp[mex], 0, sizeof(tmp[mex]));
        if (mex > L[i - 1]) memcpy(tmp[mex], tmp[mex - 1], sizeof(tmp[mex]));
      }
      for (int cnt = 0; cnt <= i; cnt++)
        if (dp[i & 1 ^ 1][mex][cnt]) {
          if (i > n) {
            update(ans, (long long)dp[i & 1 ^ 1][mex][cnt] * C(n - mex, cnt) %
                            mod * jc[cnt] % mod);
            continue;
          }
          update(dp[i & 1][mex][cnt],
                 (long long)dp[i & 1 ^ 1][mex][cnt] * (mex + cnt) % mod);
          update(dp[i & 1][mex][cnt + 1], dp[i & 1 ^ 1][mex][cnt]);
          update(tmp[mex][mex + cnt],
                 (long long)dp[i & 1 ^ 1][mex][cnt] * jc[cnt] % mod);
        }
    }
    if (i > n) break;
    for (int m = L[i]; m <= R[i]; m++)
      for (int B = 0; B <= i; B++) {
        int x = m - 1;
        if (x > R[i - 1]) x = R[i - 1];
        if (x < L[i - 1]) continue;
        int y = B + m - 1;
        if (y < 0 || y > n * 2) continue;
        update(dp[i & 1][m][B], (long long)tmp[x][y] * ivjc[B] % mod);
      }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
template <typename T>
void write(T t) {
  if (t < 0) {
    putchar('-');
    write(-t);
    return;
  }
  if (t > 9) write(t / 10);
  putchar('0' + t % 10);
}
template <typename T>
void writeln(T t) {
  write(t);
  puts("");
}
const int mod = 998244353;
const int maxn = 2010;
int n, k, jc[maxn], ivjc[maxn], iv[maxn];
int dp[2][maxn][maxn], L[maxn], R[maxn], ans, tmp[maxn][maxn * 2];
void update(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int C(int x, int y) {
  if (x < y || y < 0) return 0;
  return (long long)jc[x] * ivjc[y] % mod * ivjc[x - y] % mod;
}
int main() {
  read(n), read(k);
  iv[0] = jc[0] = ivjc[0] = iv[1] = jc[1] = ivjc[1] = 1;
  for (int i = 2; i < maxn; i++) {
    jc[i] = (long long)jc[i - 1] * i % mod;
    iv[i] = (long long)iv[mod % i] * (mod - mod / i) % mod;
    ivjc[i] = (long long)ivjc[i - 1] * iv[i] % mod;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    L[i] = max(0, x - k), R[i] = min(n + 1, x + k);
  }
  dp[0][0][0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    for (int mex = L[i]; mex <= R[i]; ++mex)
      for (int cnt = 0; cnt <= i; ++cnt) dp[i & 1][mex][cnt] = 0;
    for (int mex = L[i - 1]; mex <= R[i - 1]; ++mex) {
      if (i <= n) {
        memset(tmp[mex], 0, sizeof(tmp[mex]));
        if (mex > L[i - 1]) memcpy(tmp[mex], tmp[mex - 1], sizeof(tmp[mex]));
      }
      for (int cnt = 0; cnt <= i; cnt++)
        if (dp[i & 1 ^ 1][mex][cnt]) {
          if (i > n) {
            update(ans, (long long)dp[i & 1 ^ 1][mex][cnt] * C(n - mex, cnt) %
                            mod * jc[cnt] % mod);
            continue;
          }
          update(dp[i & 1][mex][cnt],
                 (long long)dp[i & 1 ^ 1][mex][cnt] * (mex + cnt) % mod);
          update(dp[i & 1][mex][cnt + 1], dp[i & 1 ^ 1][mex][cnt]);
          update(tmp[mex][mex + cnt],
                 (long long)dp[i & 1 ^ 1][mex][cnt] * jc[cnt] % mod);
        }
    }
    if (i > n) break;
    for (int m = L[i]; m <= R[i]; m++)
      for (int B = 0; B <= i; B++) {
        int x = m - 1;
        if (x > R[i - 1]) x = R[i - 1];
        if (x < L[i - 1]) continue;
        int y = B + m - 1;
        if (y < 0 || y > n * 2) continue;
        update(dp[i & 1][m][B], (long long)tmp[x][y] * ivjc[B] % mod);
      }
  }
  printf("%d\n", ans);
  return 0;
}

```