#include <bits/stdc++.h>
using namespace std;
long long n, k, s[1001][2001][4], g = 998244353;
int main() {
  cin >> n >> k;
  s[1][1][0] = s[1][2][1] = s[1][2][2] = s[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= i * 2; j++) {
      s[i][j][0] = (s[i - 1][j][0] + s[i - 1][j][1] + s[i - 1][j][2] +
                    s[i - 1][j - 1][3]) %
                   g;
      if (j >= 2)
        s[i][j][1] = (s[i - 1][j - 1][0] + s[i - 1][j][1] + s[i - 1][j - 2][2] +
                      s[i - 1][j - 1][3]) %
                     g;
      if (j >= 2)
        s[i][j][2] = (s[i - 1][j - 1][0] + s[i - 1][j][2] + s[i - 1][j - 2][1] +
                      s[i - 1][j - 1][3]) %
                     g;
      s[i][j][3] = (s[i - 1][j - 1][0] + s[i - 1][j][1] + s[i - 1][j][2] +
                    s[i - 1][j][3]) %
                   g;
    }
  }
  cout << (s[n][k][0] + s[n][k][1] + s[n][k][2] + s[n][k][3]) % g;
  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 n, k, s[1001][2001][4], g = 998244353;
int main() {
  cin >> n >> k;
  s[1][1][0] = s[1][2][1] = s[1][2][2] = s[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= i * 2; j++) {
      s[i][j][0] = (s[i - 1][j][0] + s[i - 1][j][1] + s[i - 1][j][2] +
                    s[i - 1][j - 1][3]) %
                   g;
      if (j >= 2)
        s[i][j][1] = (s[i - 1][j - 1][0] + s[i - 1][j][1] + s[i - 1][j - 2][2] +
                      s[i - 1][j - 1][3]) %
                     g;
      if (j >= 2)
        s[i][j][2] = (s[i - 1][j - 1][0] + s[i - 1][j][2] + s[i - 1][j - 2][1] +
                      s[i - 1][j - 1][3]) %
                     g;
      s[i][j][3] = (s[i - 1][j - 1][0] + s[i - 1][j][1] + s[i - 1][j][2] +
                    s[i - 1][j][3]) %
                   g;
    }
  }
  cout << (s[n][k][0] + s[n][k][1] + s[n][k][2] + s[n][k][3]) % g;
  return 0;
}

```