#include <bits/stdc++.h>
using namespace std;
int n, k;
long long dp[1010][2010][3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  dp[1][1][1] = dp[1][2][2] = 2;
  for (int i = 2; i <= n; ++i) {
    dp[i][1][1] = dp[i][i << 1][2] = 2;
    for (int j = 1; j < 2 * i - 1; ++j) {
      dp[i][j][1] = dp[i - 1][j - 1][1] + dp[i - 1][j][1] + dp[i - 1][j][2] * 2;
      dp[i][j][1] %= 998244353;
      dp[i][j][2] =
          dp[i - 1][j - 2][2] + dp[i - 1][j - 1][1] * 2 + dp[i - 1][j][2];
      dp[i][j][2] %= 998244353;
    }
  }
  cout << (dp[n][k][1] + dp[n][k][2]) % 998244353;
  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;
int n, k;
long long dp[1010][2010][3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  dp[1][1][1] = dp[1][2][2] = 2;
  for (int i = 2; i <= n; ++i) {
    dp[i][1][1] = dp[i][i << 1][2] = 2;
    for (int j = 1; j < 2 * i - 1; ++j) {
      dp[i][j][1] = dp[i - 1][j - 1][1] + dp[i - 1][j][1] + dp[i - 1][j][2] * 2;
      dp[i][j][1] %= 998244353;
      dp[i][j][2] =
          dp[i - 1][j - 2][2] + dp[i - 1][j - 1][1] * 2 + dp[i - 1][j][2];
      dp[i][j][2] %= 998244353;
    }
  }
  cout << (dp[n][k][1] + dp[n][k][2]) % 998244353;
  return 0;
}

```