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

### Prompt

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

```