#include <bits/stdc++.h>
using namespace std;
long long n, m, i, j, k, mm;
long long f[2001][3001][2] = {0};
int main() {
  cin >> n >> m;
  f[1][1][1] = 2;
  f[1][2][0] = 2;
  for (i = 2; i <= n; ++i) {
    if (2 * i < m)
      mm = 2 * i;
    else
      mm = m;
    for (j = 1; j <= mm; ++j) {
      f[i][j][1] = (f[i - 1][j][0] * 2 + f[i - 1][j][1] + f[i - 1][j - 1][1]) %
                   998244353;
      f[i][j][0] =
          (f[i - 1][j][0] + f[i - 1][j - 1][1] * 2 + f[i - 1][j - 2][0]) %
          998244353;
    }
  }
  cout << (f[n][m][0] + f[n][m][1]) % 998244353;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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, m, i, j, k, mm;
long long f[2001][3001][2] = {0};
int main() {
  cin >> n >> m;
  f[1][1][1] = 2;
  f[1][2][0] = 2;
  for (i = 2; i <= n; ++i) {
    if (2 * i < m)
      mm = 2 * i;
    else
      mm = m;
    for (j = 1; j <= mm; ++j) {
      f[i][j][1] = (f[i - 1][j][0] * 2 + f[i - 1][j][1] + f[i - 1][j - 1][1]) %
                   998244353;
      f[i][j][0] =
          (f[i - 1][j][0] + f[i - 1][j - 1][1] * 2 + f[i - 1][j - 2][0]) %
          998244353;
    }
  }
  cout << (f[n][m][0] + f[n][m][1]) % 998244353;
  return 0;
}

```