#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
const int mod = 998244353;
long long dp[maxn][maxn][4];
int main() {
  int n, m, k;
  cin >> n >> k;
  if (k == 2 * n) {
    cout << "2" << endl;
    return 0;
  } else if (k == 2 * n - 1 && n != 1) {
    cout << "0" << endl;
    return 0;
  }
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][0] = 1;
  dp[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j][0] = dp[i - 1][j - 1][3] + dp[i - 1][j][0] + dp[i - 1][j][1] +
                    dp[i - 1][j][2];
      dp[i][j][1] = dp[i - 1][j - 1][3] + dp[i - 1][j - 1][0] +
                    dp[i - 1][j][1] + (j >= 2 ? dp[i - 1][j - 2][2] : 0);
      dp[i][j][2] = dp[i - 1][j - 1][3] + dp[i - 1][j - 1][0] +
                    (j > 2 ? dp[i - 1][j - 2][1] : 0) + dp[i - 1][j][2];
      dp[i][j][3] = dp[i - 1][j][3] + dp[i - 1][j - 1][0] + dp[i - 1][j][1] +
                    dp[i - 1][j][2];
      for (m = 0; m < 4; m++) dp[i][j][m] %= 998244353;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= 3; i++) {
    ans = (ans + dp[n][k][i]) % mod;
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution 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;
const int maxn = 1e3 + 5;
const int mod = 998244353;
long long dp[maxn][maxn][4];
int main() {
  int n, m, k;
  cin >> n >> k;
  if (k == 2 * n) {
    cout << "2" << endl;
    return 0;
  } else if (k == 2 * n - 1 && n != 1) {
    cout << "0" << endl;
    return 0;
  }
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][0] = 1;
  dp[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j][0] = dp[i - 1][j - 1][3] + dp[i - 1][j][0] + dp[i - 1][j][1] +
                    dp[i - 1][j][2];
      dp[i][j][1] = dp[i - 1][j - 1][3] + dp[i - 1][j - 1][0] +
                    dp[i - 1][j][1] + (j >= 2 ? dp[i - 1][j - 2][2] : 0);
      dp[i][j][2] = dp[i - 1][j - 1][3] + dp[i - 1][j - 1][0] +
                    (j > 2 ? dp[i - 1][j - 2][1] : 0) + dp[i - 1][j][2];
      dp[i][j][3] = dp[i - 1][j][3] + dp[i - 1][j - 1][0] + dp[i - 1][j][1] +
                    dp[i - 1][j][2];
      for (m = 0; m < 4; m++) dp[i][j][m] %= 998244353;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= 3; i++) {
    ans = (ans + dp[n][k][i]) % mod;
  }
  cout << ans << endl;
}

```