#include <bits/stdc++.h>
int n, k, k2;
long long ans = 0;
int b[2002];
long long fac[2052], inv[2052];
long long f[2][2002][102];
struct aaa {
  int x, y, z;
};
inline int max(int x, int y) { return x > y ? x : y; }
inline long long Pow(long long a, int b) {
  if (!b) return 1;
  if (b == 1) return a;
  long long c = Pow(a, (b >> 1));
  c = (c * c) % 998244353;
  if (b & 1) c = (c * a) % 998244353;
  return c;
}
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= n + k; ++i) fac[i] = (fac[i - 1] * i) % 998244353;
  inv[n + k] = Pow(fac[n + k], 998244353 - 2);
  for (int i = n + k; i; --i) inv[i - 1] = (inv[i] * i) % 998244353;
}
inline long long C(int a, int b) {
  if (b < 0 || a < b) return 0;
  return (((fac[a] * inv[b]) % 998244353) * inv[a - b]) % 998244353;
}
inline bool check(int x) { return (x >= 0 && x <= k2); }
int main() {
  scanf("%d%d", &n, &k), k2 = (k << 1), b[0] = 0, f[0][0][k] = 1, init();
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  for (int i = 1, i0 = 0, i1 = 1; i <= n; ++i, i0 ^= 1, i1 ^= 1) {
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) f[i1][j][o] = 0;
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) {
        if (check(max(b[i - 1] + o + 1 - b[i], 0)))
          (f[i1][j + 1][max(b[i - 1] + o + 1 - b[i], 0)] += f[i0][j][o]) %=
              998244353;
        if (check(b[i - 1] + o + j + 2 - b[i]))
          (f[i1][j + 1][b[i - 1] + o + j + 2 - b[i]] -= f[i0][j][o]) %=
              998244353;
      }
    for (int j = 0; j <= i; ++j)
      for (int o = 1; o <= k2; ++o)
        (f[i1][j][o] += f[i1][j][o - 1]) %= 998244353;
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) {
        if (check(b[i - 1] + o - b[i]))
          (f[i1][j][b[i - 1] + o - b[i]] += f[i0][j][o] * j) %= 998244353;
        if (check(b[i - 1] + o - b[i]))
          (f[i1][j + 1][b[i - 1] + o - b[i]] +=
           f[i0][j][o] * (j - b[i - 1] - o + k + 1)) %= 998244353;
      }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= k2; ++j)
      (ans += f[n & 1][i][j] * C(n - b[n] - j + k, i - b[n] - j + k)) %=
          998244353;
  return 0 & printf("%lld", (ans + 998244353) % 998244353);
}

### Prompt

Please formulate 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>
int n, k, k2;
long long ans = 0;
int b[2002];
long long fac[2052], inv[2052];
long long f[2][2002][102];
struct aaa {
  int x, y, z;
};
inline int max(int x, int y) { return x > y ? x : y; }
inline long long Pow(long long a, int b) {
  if (!b) return 1;
  if (b == 1) return a;
  long long c = Pow(a, (b >> 1));
  c = (c * c) % 998244353;
  if (b & 1) c = (c * a) % 998244353;
  return c;
}
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= n + k; ++i) fac[i] = (fac[i - 1] * i) % 998244353;
  inv[n + k] = Pow(fac[n + k], 998244353 - 2);
  for (int i = n + k; i; --i) inv[i - 1] = (inv[i] * i) % 998244353;
}
inline long long C(int a, int b) {
  if (b < 0 || a < b) return 0;
  return (((fac[a] * inv[b]) % 998244353) * inv[a - b]) % 998244353;
}
inline bool check(int x) { return (x >= 0 && x <= k2); }
int main() {
  scanf("%d%d", &n, &k), k2 = (k << 1), b[0] = 0, f[0][0][k] = 1, init();
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  for (int i = 1, i0 = 0, i1 = 1; i <= n; ++i, i0 ^= 1, i1 ^= 1) {
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) f[i1][j][o] = 0;
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) {
        if (check(max(b[i - 1] + o + 1 - b[i], 0)))
          (f[i1][j + 1][max(b[i - 1] + o + 1 - b[i], 0)] += f[i0][j][o]) %=
              998244353;
        if (check(b[i - 1] + o + j + 2 - b[i]))
          (f[i1][j + 1][b[i - 1] + o + j + 2 - b[i]] -= f[i0][j][o]) %=
              998244353;
      }
    for (int j = 0; j <= i; ++j)
      for (int o = 1; o <= k2; ++o)
        (f[i1][j][o] += f[i1][j][o - 1]) %= 998244353;
    for (int j = 0; j <= i; ++j)
      for (int o = 0; o <= k2; ++o) {
        if (check(b[i - 1] + o - b[i]))
          (f[i1][j][b[i - 1] + o - b[i]] += f[i0][j][o] * j) %= 998244353;
        if (check(b[i - 1] + o - b[i]))
          (f[i1][j + 1][b[i - 1] + o - b[i]] +=
           f[i0][j][o] * (j - b[i - 1] - o + k + 1)) %= 998244353;
      }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= k2; ++j)
      (ans += f[n & 1][i][j] * C(n - b[n] - j + k, i - b[n] - j + k)) %=
          998244353;
  return 0 & printf("%lld", (ans + 998244353) % 998244353);
}

```