#include <bits/stdc++.h>
using namespace std;
const int N = 4444;
const int M = 111;
const int MO = 998244353;
void ad(int& x, int y) {
  x += y;
  if (x >= MO) x -= MO;
}
int mul(int x, int y) { return (long long)x * y % MO; }
int fpow(int x, int y = MO - 2) {
  int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
int fac(int o) {
  int i, x = 1;
  for (i = 1; i <= o; i++) x = mul(x, i);
  return x;
}
int n, m, b[N], c[N][N], f[2][M][N];
bool ok(int x) { return x >= 0 && x <= m; }
int main() {
  int i, j, k, o, s, x;
  n = 4022;
  for (i = 0; i <= n; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MO;
  }
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", b + i), b[i] -= m;
  b[0] = 0;
  m = m * 2;
  f[0][0][0] = 1;
  o = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        if (!f[o][j][k]) continue;
        if (ok(j + b[i - 1] - b[i] + 1))
          ad(f[o ^ 1][j + b[i - 1] - b[i] + 1][k], f[o][j][k]);
        else {
          if (j + b[i - 1] - b[i] + 1 < 0) {
            x = j + b[i - 1] - b[i] + 1;
            if (k + x >= 0)
              ad(f[o ^ 1][j + b[i - 1] - b[i] + 1 - x][k + x], f[o][j][k]);
          }
        }
      }
    }
    for (j = 1; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        ad(f[o ^ 1][j][k], f[o ^ 1][j - 1][k + 1]);
      }
    }
    for (j = 0; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        if (!f[o][j][k]) continue;
        if (ok(j + b[i - 1] - b[i])) {
          ad(f[o ^ 1][j + b[i - 1] - b[i]][k],
             mul(f[o][j][k], j + b[i - 1] + k));
          ad(f[o ^ 1][j + b[i - 1] - b[i]][k + 1], mul(f[o][j][k], k + 1));
        }
      }
    }
    for (j = 0; j < M; j++)
      for (k = 0; k < N; k++) f[o][j][k] = 0;
    o ^= 1;
  }
  s = 0;
  for (j = 0; j <= m; j++)
    for (k = 0; k <= n; k++)
      if (b[n] + j >= 0 && b[n] + j <= n)
        ad(s, mul(f[o][j][k], c[n - (b[n] + j)][k]));
  printf("%d\n", s);
  return 0;
}

### 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>
using namespace std;
const int N = 4444;
const int M = 111;
const int MO = 998244353;
void ad(int& x, int y) {
  x += y;
  if (x >= MO) x -= MO;
}
int mul(int x, int y) { return (long long)x * y % MO; }
int fpow(int x, int y = MO - 2) {
  int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
int fac(int o) {
  int i, x = 1;
  for (i = 1; i <= o; i++) x = mul(x, i);
  return x;
}
int n, m, b[N], c[N][N], f[2][M][N];
bool ok(int x) { return x >= 0 && x <= m; }
int main() {
  int i, j, k, o, s, x;
  n = 4022;
  for (i = 0; i <= n; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MO;
  }
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", b + i), b[i] -= m;
  b[0] = 0;
  m = m * 2;
  f[0][0][0] = 1;
  o = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        if (!f[o][j][k]) continue;
        if (ok(j + b[i - 1] - b[i] + 1))
          ad(f[o ^ 1][j + b[i - 1] - b[i] + 1][k], f[o][j][k]);
        else {
          if (j + b[i - 1] - b[i] + 1 < 0) {
            x = j + b[i - 1] - b[i] + 1;
            if (k + x >= 0)
              ad(f[o ^ 1][j + b[i - 1] - b[i] + 1 - x][k + x], f[o][j][k]);
          }
        }
      }
    }
    for (j = 1; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        ad(f[o ^ 1][j][k], f[o ^ 1][j - 1][k + 1]);
      }
    }
    for (j = 0; j <= m; j++) {
      for (k = 0; k <= i; k++) {
        if (!f[o][j][k]) continue;
        if (ok(j + b[i - 1] - b[i])) {
          ad(f[o ^ 1][j + b[i - 1] - b[i]][k],
             mul(f[o][j][k], j + b[i - 1] + k));
          ad(f[o ^ 1][j + b[i - 1] - b[i]][k + 1], mul(f[o][j][k], k + 1));
        }
      }
    }
    for (j = 0; j < M; j++)
      for (k = 0; k < N; k++) f[o][j][k] = 0;
    o ^= 1;
  }
  s = 0;
  for (j = 0; j <= m; j++)
    for (k = 0; k <= n; k++)
      if (b[n] + j >= 0 && b[n] + j <= n)
        ad(s, mul(f[o][j][k], c[n - (b[n] + j)][k]));
  printf("%d\n", s);
  return 0;
}

```