#include <bits/stdc++.h>
using namespace std;
long long t[2000][1000][2] = {};
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) t[0][i][0] = 2;
  t[1][0][1] = 2;
  for (int i = 1; i < k; i++) {
    for (int j = 1; j < n; j++) {
      t[i][j][0] =
          ((t[i][j - 1][0] + t[i][j - 1][1] * 2 % 998244353) % 998244353 +
           t[i - 1][j - 1][0]) %
          998244353;
      t[i][j][1] =
          ((t[i][j - 1][1] + t[i - 1][j - 1][0] * 2 % 998244353) % 998244353 +
           (i > 1 ? t[i - 2][j - 1][1] : 0)) %
          998244353;
    }
  }
  cout << (t[k - 1][n - 1][0] + t[k - 1][n - 1][1]) % 998244353 << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long t[2000][1000][2] = {};
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) t[0][i][0] = 2;
  t[1][0][1] = 2;
  for (int i = 1; i < k; i++) {
    for (int j = 1; j < n; j++) {
      t[i][j][0] =
          ((t[i][j - 1][0] + t[i][j - 1][1] * 2 % 998244353) % 998244353 +
           t[i - 1][j - 1][0]) %
          998244353;
      t[i][j][1] =
          ((t[i][j - 1][1] + t[i - 1][j - 1][0] * 2 % 998244353) % 998244353 +
           (i > 1 ? t[i - 2][j - 1][1] : 0)) %
          998244353;
    }
  }
  cout << (t[k - 1][n - 1][0] + t[k - 1][n - 1][1]) % 998244353 << "\n";
  return 0;
}

```