#include <bits/stdc++.h>
using namespace std;
const int P = 998244353;
int dp[103][2003], tp[103][2003], R[2003], TR[2003], N, K, A[2003];
int upd(int x) { return x + (x >> 31 & P); }
int add(int x, int y) { return upd(x + y - P); }
int sub(int x, int y) { return upd(x - y); }
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; ++i) cin >> A[i];
  for (int i = A[N] - K; i <= A[N] + K; ++i)
    if (i >= 0 && i <= N) dp[i + K - A[N]][1] = 1;
  A[0] = -K;
  for (int i = N; i; --i) {
    memset(tp, 0, sizeof(tp));
    for (int j = 0; j <= 2 * K; ++j)
      if (A[i] + j - K >= A[i - 1] - K && A[i] + j - K <= A[i - 1] + K) {
        int pos = A[i] + j - K - A[i - 1] + K;
        for (int l = 0; l <= N + 1; ++l)
          if (dp[j][l])
            tp[pos][l] = (tp[pos][l] + 1ll * dp[j][l] * (N + 1 - l)) % P;
      }
    memset(R, 0, sizeof(R));
    for (int t = A[i] + K - 1; t >= A[i - 1] - K; --t) {
      for (int j = N + 1; ~j; --j)
        if (R[j]) R[j + 1] = sub(R[j + 1], R[j]);
      if (t + 1 >= A[i] - K)
        for (int j = 0; j <= N + 1; ++j)
          R[j] = add(R[j], dp[t + 1 - A[i] + K][j]);
      if (t <= A[i - 1] + K)
        for (int j = 0; j <= N + 1; ++j)
          tp[t - A[i - 1] + K][j + 1] = add(tp[t - A[i - 1] + K][j + 1], R[j]);
    }
    memcpy(dp, tp, sizeof(dp));
  }
  int sum = 0;
  for (int i = 0; i <= N + 1; ++i) sum = (sum + dp[2 * K][i]) % P;
  cout << sum;
  return 0;
}

### Prompt

Your challenge is to write 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 P = 998244353;
int dp[103][2003], tp[103][2003], R[2003], TR[2003], N, K, A[2003];
int upd(int x) { return x + (x >> 31 & P); }
int add(int x, int y) { return upd(x + y - P); }
int sub(int x, int y) { return upd(x - y); }
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; ++i) cin >> A[i];
  for (int i = A[N] - K; i <= A[N] + K; ++i)
    if (i >= 0 && i <= N) dp[i + K - A[N]][1] = 1;
  A[0] = -K;
  for (int i = N; i; --i) {
    memset(tp, 0, sizeof(tp));
    for (int j = 0; j <= 2 * K; ++j)
      if (A[i] + j - K >= A[i - 1] - K && A[i] + j - K <= A[i - 1] + K) {
        int pos = A[i] + j - K - A[i - 1] + K;
        for (int l = 0; l <= N + 1; ++l)
          if (dp[j][l])
            tp[pos][l] = (tp[pos][l] + 1ll * dp[j][l] * (N + 1 - l)) % P;
      }
    memset(R, 0, sizeof(R));
    for (int t = A[i] + K - 1; t >= A[i - 1] - K; --t) {
      for (int j = N + 1; ~j; --j)
        if (R[j]) R[j + 1] = sub(R[j + 1], R[j]);
      if (t + 1 >= A[i] - K)
        for (int j = 0; j <= N + 1; ++j)
          R[j] = add(R[j], dp[t + 1 - A[i] + K][j]);
      if (t <= A[i - 1] + K)
        for (int j = 0; j <= N + 1; ++j)
          tp[t - A[i - 1] + K][j + 1] = add(tp[t - A[i - 1] + K][j + 1], R[j]);
    }
    memcpy(dp, tp, sizeof(dp));
  }
  int sum = 0;
  for (int i = 0; i <= N + 1; ++i) sum = (sum + dp[2 * K][i]) % P;
  cout << sum;
  return 0;
}

```