#include <bits/stdc++.h>
using namespace std;
int n, k;
int d[1005][2 * 1005][5];
int main() {
  cin >> n >> k;
  d[1][1][1] = d[1][1][4] = 1;
  d[1][2][2] = d[1][2][3] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 2 * i; j++) {
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][2] = (d[i + 1][j + 1][2] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][3] = (d[i + 1][j + 1][3] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][4] = (d[i + 1][j + 1][4] + d[i][j][1]) % 998244353;
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][2]) % 998244353;
      d[i + 1][j][2] = (d[i + 1][j][2] + d[i][j][2]) % 998244353;
      d[i + 1][j + 2][3] = (d[i + 1][j + 2][3] + d[i][j][2]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][2]) % 998244353;
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][3]) % 998244353;
      d[i + 1][j + 2][2] = (d[i + 1][j + 2][2] + d[i][j][3]) % 998244353;
      d[i + 1][j][3] = (d[i + 1][j][3] + d[i][j][3]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][3]) % 998244353;
      d[i + 1][j + 1][1] = (d[i + 1][j + 1][1] + d[i][j][4]) % 998244353;
      d[i + 1][j + 1][2] = (d[i + 1][j + 1][2] + d[i][j][4]) % 998244353;
      d[i + 1][j + 1][3] = (d[i + 1][j + 1][3] + d[i][j][4]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][4]) % 998244353;
    }
  }
  cout << ((d[n][k][1] + d[n][k][2]) % 998244353 +
           (d[n][k][3] + d[n][k][4]) % 998244353) %
              998244353;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You are given a grid, consisting of 2 rows and n columns. Each cell of this grid should be colored either black or white.

Two cells are considered neighbours if they have a common border and share the same color. Two cells A and B belong to the same component if they are neighbours, or if there is a neighbour of A that belongs to the same component with B.

Let's call some bicoloring beautiful if it has exactly k components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo 998244353.

Input

The only line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ 2n) — the number of columns in a grid and the number of components required.

Output

Print a single integer — the number of beautiful bicolorings modulo 998244353.

Examples

Input

3 4


Output

12


Input

4 1


Output

2


Input

1 2


Output

2

Note

One of possible bicolorings in sample 1:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, k;
int d[1005][2 * 1005][5];
int main() {
  cin >> n >> k;
  d[1][1][1] = d[1][1][4] = 1;
  d[1][2][2] = d[1][2][3] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 2 * i; j++) {
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][2] = (d[i + 1][j + 1][2] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][3] = (d[i + 1][j + 1][3] + d[i][j][1]) % 998244353;
      d[i + 1][j + 1][4] = (d[i + 1][j + 1][4] + d[i][j][1]) % 998244353;
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][2]) % 998244353;
      d[i + 1][j][2] = (d[i + 1][j][2] + d[i][j][2]) % 998244353;
      d[i + 1][j + 2][3] = (d[i + 1][j + 2][3] + d[i][j][2]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][2]) % 998244353;
      d[i + 1][j][1] = (d[i + 1][j][1] + d[i][j][3]) % 998244353;
      d[i + 1][j + 2][2] = (d[i + 1][j + 2][2] + d[i][j][3]) % 998244353;
      d[i + 1][j][3] = (d[i + 1][j][3] + d[i][j][3]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][3]) % 998244353;
      d[i + 1][j + 1][1] = (d[i + 1][j + 1][1] + d[i][j][4]) % 998244353;
      d[i + 1][j + 1][2] = (d[i + 1][j + 1][2] + d[i][j][4]) % 998244353;
      d[i + 1][j + 1][3] = (d[i + 1][j + 1][3] + d[i][j][4]) % 998244353;
      d[i + 1][j][4] = (d[i + 1][j][4] + d[i][j][4]) % 998244353;
    }
  }
  cout << ((d[n][k][1] + d[n][k][2]) % 998244353 +
           (d[n][k][3] + d[n][k][4]) % 998244353) %
              998244353;
}

```