#include <bits/stdc++.h>
using namespace std;
struct Mat {
  int same, dif;
} mat[2 * 1002][1002];
int n, k;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) mat[1][i].same = 2;
  mat[2][1].same = 0;
  mat[2][1].dif = 2;
  for (int i = 2; i <= k; i++) {
    for (int j = max((i / 2) + i % 2, 2); j <= n; j++) {
      mat[i][j].same =
          (mat[i][j - 1].same + mat[i][j - 1].dif * 2LL % 998244353 +
           mat[i - 1][j - 1].same) %
          998244353;
      mat[i][j].dif = (mat[i - 1][j - 1].same * 2LL % 998244353 +
                       mat[i][j - 1].dif + mat[i - 2][j - 1].dif) %
                      998244353;
    }
  }
  cout << (mat[k][n].same + mat[k][n].dif) % 998244353;
}

### Prompt

In CPP, your task is to solve 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;
struct Mat {
  int same, dif;
} mat[2 * 1002][1002];
int n, k;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) mat[1][i].same = 2;
  mat[2][1].same = 0;
  mat[2][1].dif = 2;
  for (int i = 2; i <= k; i++) {
    for (int j = max((i / 2) + i % 2, 2); j <= n; j++) {
      mat[i][j].same =
          (mat[i][j - 1].same + mat[i][j - 1].dif * 2LL % 998244353 +
           mat[i - 1][j - 1].same) %
          998244353;
      mat[i][j].dif = (mat[i - 1][j - 1].same * 2LL % 998244353 +
                       mat[i][j - 1].dif + mat[i - 2][j - 1].dif) %
                      998244353;
    }
  }
  cout << (mat[k][n].same + mat[k][n].dif) % 998244353;
}

```