#include <bits/stdc++.h>
using namespace std;
long long dp[1001][2001][2][2];
void solve() {
  int n, k;
  cin >> n >> k;
  dp[1][2][0][1] = 1;
  dp[1][2][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][1][1][1] = 1;
    dp[i][1][0][0] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= k; j++) {
      dp[i][j][1][1] = dp[i - 1][j][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j - 1][0][0];
      dp[i][j][0][0] = dp[i - 1][j - 1][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j][0][0];
      dp[i][j][1][0] = dp[i - 1][j - 1][1][1] + dp[i - 1][j - 2][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j - 1][0][0];
      dp[i][j][0][1] = dp[i - 1][j - 1][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j - 2][1][0] + dp[i - 1][j - 1][0][0];
      for (int x = 0; x < 2; x++) {
        for (int y = 0; y < 2; y++) {
          dp[i][j][x][y] %= 998244353;
        }
      }
      cout << "\n";
    }
  }
  long long ans = 0;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      ans += dp[n][k][i][j];
    }
  }
  cout << ans % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution 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 dp[1001][2001][2][2];
void solve() {
  int n, k;
  cin >> n >> k;
  dp[1][2][0][1] = 1;
  dp[1][2][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][1][1][1] = 1;
    dp[i][1][0][0] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= k; j++) {
      dp[i][j][1][1] = dp[i - 1][j][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j - 1][0][0];
      dp[i][j][0][0] = dp[i - 1][j - 1][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j][0][0];
      dp[i][j][1][0] = dp[i - 1][j - 1][1][1] + dp[i - 1][j - 2][0][1] +
                       dp[i - 1][j][1][0] + dp[i - 1][j - 1][0][0];
      dp[i][j][0][1] = dp[i - 1][j - 1][1][1] + dp[i - 1][j][0][1] +
                       dp[i - 1][j - 2][1][0] + dp[i - 1][j - 1][0][0];
      for (int x = 0; x < 2; x++) {
        for (int y = 0; y < 2; y++) {
          dp[i][j][x][y] %= 998244353;
        }
      }
      cout << "\n";
    }
  }
  long long ans = 0;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      ans += dp[n][k][i][j];
    }
  }
  cout << ans % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}

```