#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
int n, k;
int b[2010];
int dp[2][110][2010];
int D[5310];
int fac[4010], ifac[4010];
const int mod = 998244353;
inline int qpow(int x, int k = mod - 2) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) b[i] = read();
  dp[0][k][0] = 1;
  bool fl = 0;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n]);
  for (int i = n - 1; i >= 0; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; ++i) {
    fl ^= 1;
    memset(dp[fl], 0, sizeof(dp[fl]));
    memset(D, 0, sizeof(D));
    int it = -min(k, b[i - 1]);
    for (int j = -min(b[i], k); j <= k; ++j) {
      int val = b[i] + j;
      while (it < val - b[i - 1] && it <= k) {
        for (int l = 0; l <= i; ++l)
          (D[l + b[i - 1] + it] +=
           1ll * dp[fl ^ 1][it + k][l] * fac[l] % mod) %= mod;
        ++it;
      }
      for (int l = 0; l <= i; ++l) {
        if (l + val > 0)
          dp[fl][j + k][l] = 1ll * D[l + val - 1] * ifac[l] % mod;
        if (abs(val - b[i - 1]) <= k) {
          (dp[fl][j + k][l] +=
           1ll * (val + l) * dp[fl ^ 1][k + val - b[i - 1]][l] % mod) %= mod;
          if (l)
            (dp[fl][j + k][l] += dp[fl ^ 1][k + val - b[i - 1]][l - 1]) %= mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = -min(b[n], k); i <= k && i + b[n] <= n; ++i) {
    for (int l = 0; l <= n && b[n] + i + l <= n; ++l) {
      (ans += 1ll * dp[fl][i + k][l] * fac[n - (b[n] + i)] % mod *
              ifac[n - (b[n] + i + l)] % mod) %= mod;
    }
  }
  printf("%d\n", ans);
}

### 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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
int n, k;
int b[2010];
int dp[2][110][2010];
int D[5310];
int fac[4010], ifac[4010];
const int mod = 998244353;
inline int qpow(int x, int k = mod - 2) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) b[i] = read();
  dp[0][k][0] = 1;
  bool fl = 0;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n]);
  for (int i = n - 1; i >= 0; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; ++i) {
    fl ^= 1;
    memset(dp[fl], 0, sizeof(dp[fl]));
    memset(D, 0, sizeof(D));
    int it = -min(k, b[i - 1]);
    for (int j = -min(b[i], k); j <= k; ++j) {
      int val = b[i] + j;
      while (it < val - b[i - 1] && it <= k) {
        for (int l = 0; l <= i; ++l)
          (D[l + b[i - 1] + it] +=
           1ll * dp[fl ^ 1][it + k][l] * fac[l] % mod) %= mod;
        ++it;
      }
      for (int l = 0; l <= i; ++l) {
        if (l + val > 0)
          dp[fl][j + k][l] = 1ll * D[l + val - 1] * ifac[l] % mod;
        if (abs(val - b[i - 1]) <= k) {
          (dp[fl][j + k][l] +=
           1ll * (val + l) * dp[fl ^ 1][k + val - b[i - 1]][l] % mod) %= mod;
          if (l)
            (dp[fl][j + k][l] += dp[fl ^ 1][k + val - b[i - 1]][l - 1]) %= mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = -min(b[n], k); i <= k && i + b[n] <= n; ++i) {
    for (int l = 0; l <= n && b[n] + i + l <= n; ++l) {
      (ans += 1ll * dp[fl][i + k][l] * fac[n - (b[n] + i)] % mod *
              ifac[n - (b[n] + i + l)] % mod) %= mod;
    }
  }
  printf("%d\n", ans);
}

```