#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;
int Bit(int mask, int b) { return (mask >> b) & 1; }
template <class T>
bool ckmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool ckmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
const int INFi = 2e9;
const ll INF = 2e18;
const int maxN = 2e5;
const int LG = 30;
const int md = 998244353;
inline int add(const int &a, const int &b) {
  return a + b >= md ? a + b - md : a + b;
}
inline int sub(const int &a, const int &b) {
  return a - b < 0 ? a - b + md : a - b;
}
inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
int binpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) res = mul(res, a);
  return res;
}
int rev(int a) { return binpow(a, md - 2); }
const int maxF = 2e5 + 5;
int fact[maxF];
int rfact[maxF];
void init() {
  fact[0] = 1;
  for (int i = 1; i < maxF; ++i) fact[i] = mul(fact[i - 1], i);
  rfact[maxF - 1] = rev(fact[maxF - 1]);
  for (int i = maxF - 1; i >= 1; --i) {
    rfact[i - 1] = mul(rfact[i], i);
  }
}
int C(int n, int k) {
  if (k < 0 || n < k) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
void add2(int &a, int b) { a = add(a, b); }
void solve() {
  init();
  int n;
  cin >> n;
  int k;
  cin >> k;
  vi b(n);
  for (int i = 0; i < (n); ++i) cin >> b[i];
  vi l(n + 1), r(n + 1);
  for (int i = 0; i < (n); ++i) l[i + 1] = b[i] - k;
  for (int i = 0; i < (n); ++i) r[i + 1] = b[i] + k;
  for (int i = 1; i <= n; ++i) {
    l[i] = max(l[i - 1], l[i]);
    r[i] = min(r[i], i);
    if (l[i] > r[i]) {
      cout << "0\n";
      return;
    }
  }
  vvi dp(1, vi(n + 1, 0));
  dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    vvi dp2(r[i] - l[i] + 1, vi(n + 1, 0));
    vvi dp3(r[i] - l[i] + 1, vi(n + 1, 0));
    for (int was = l[i - 1]; was <= r[i - 1]; ++was) {
      int t = was - l[i - 1];
      for (int cnt = 0; cnt < i; ++cnt) {
        if (dp[t][cnt] == 0) continue;
        if (l[i] <= was && was <= r[i]) {
          add2(dp2[was - l[i]][cnt], mul(dp[t][cnt], was + cnt));
          add2(dp2[was - l[i]][cnt + 1], mul(cnt + 1, dp[t][cnt]));
        }
        for (int nxt = max(was + 1, l[i]); nxt <= r[i]; ++nxt) {
          if (cnt < nxt - was - 1) break;
          add2(dp3[nxt - l[i]][cnt - (nxt - was - 1)], dp[t][cnt]);
          break;
        }
      }
    }
    for (int j = 0; j < (r[i] - l[i]); ++j) {
      for (int t = n; t >= 1; --t) {
        dp3[j + 1][t - 1] = add(dp3[j + 1][t - 1], dp3[j][t]);
      }
    }
    for (int j = 0; j < (r[i] - l[i] + 1); ++j)
      for (int t = 0; t < (n + 1); ++t) dp2[j][t] = add(dp2[j][t], dp3[j][t]);
    swap(dp, dp2);
  }
  int ans = 0;
  for (int to = 0; to <= r[n] - l[n]; ++to) {
    for (int cnt = 0; cnt < (n + 1); ++cnt) {
      if (dp[to][cnt] == 0) continue;
      ans = add(ans, mul(dp[to][cnt], C(n - (to + l[n]), cnt)));
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 0; i < (t); ++i) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;
int Bit(int mask, int b) { return (mask >> b) & 1; }
template <class T>
bool ckmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool ckmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
const int INFi = 2e9;
const ll INF = 2e18;
const int maxN = 2e5;
const int LG = 30;
const int md = 998244353;
inline int add(const int &a, const int &b) {
  return a + b >= md ? a + b - md : a + b;
}
inline int sub(const int &a, const int &b) {
  return a - b < 0 ? a - b + md : a - b;
}
inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
int binpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) res = mul(res, a);
  return res;
}
int rev(int a) { return binpow(a, md - 2); }
const int maxF = 2e5 + 5;
int fact[maxF];
int rfact[maxF];
void init() {
  fact[0] = 1;
  for (int i = 1; i < maxF; ++i) fact[i] = mul(fact[i - 1], i);
  rfact[maxF - 1] = rev(fact[maxF - 1]);
  for (int i = maxF - 1; i >= 1; --i) {
    rfact[i - 1] = mul(rfact[i], i);
  }
}
int C(int n, int k) {
  if (k < 0 || n < k) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
void add2(int &a, int b) { a = add(a, b); }
void solve() {
  init();
  int n;
  cin >> n;
  int k;
  cin >> k;
  vi b(n);
  for (int i = 0; i < (n); ++i) cin >> b[i];
  vi l(n + 1), r(n + 1);
  for (int i = 0; i < (n); ++i) l[i + 1] = b[i] - k;
  for (int i = 0; i < (n); ++i) r[i + 1] = b[i] + k;
  for (int i = 1; i <= n; ++i) {
    l[i] = max(l[i - 1], l[i]);
    r[i] = min(r[i], i);
    if (l[i] > r[i]) {
      cout << "0\n";
      return;
    }
  }
  vvi dp(1, vi(n + 1, 0));
  dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    vvi dp2(r[i] - l[i] + 1, vi(n + 1, 0));
    vvi dp3(r[i] - l[i] + 1, vi(n + 1, 0));
    for (int was = l[i - 1]; was <= r[i - 1]; ++was) {
      int t = was - l[i - 1];
      for (int cnt = 0; cnt < i; ++cnt) {
        if (dp[t][cnt] == 0) continue;
        if (l[i] <= was && was <= r[i]) {
          add2(dp2[was - l[i]][cnt], mul(dp[t][cnt], was + cnt));
          add2(dp2[was - l[i]][cnt + 1], mul(cnt + 1, dp[t][cnt]));
        }
        for (int nxt = max(was + 1, l[i]); nxt <= r[i]; ++nxt) {
          if (cnt < nxt - was - 1) break;
          add2(dp3[nxt - l[i]][cnt - (nxt - was - 1)], dp[t][cnt]);
          break;
        }
      }
    }
    for (int j = 0; j < (r[i] - l[i]); ++j) {
      for (int t = n; t >= 1; --t) {
        dp3[j + 1][t - 1] = add(dp3[j + 1][t - 1], dp3[j][t]);
      }
    }
    for (int j = 0; j < (r[i] - l[i] + 1); ++j)
      for (int t = 0; t < (n + 1); ++t) dp2[j][t] = add(dp2[j][t], dp3[j][t]);
    swap(dp, dp2);
  }
  int ans = 0;
  for (int to = 0; to <= r[n] - l[n]; ++to) {
    for (int cnt = 0; cnt < (n + 1); ++cnt) {
      if (dp[to][cnt] == 0) continue;
      ans = add(ans, mul(dp[to][cnt], C(n - (to + l[n]), cnt)));
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  for (int i = 0; i < (t); ++i) {
    solve();
  }
  return 0;
}

```