#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
long long dp[1001][4][4001];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  dp[1][0][1] = 1;
  dp[1][1][2] = 1;
  dp[1][2][2] = 1;
  dp[1][3][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      dp[i + 1][0][j] =
          (dp[i + 1][0][j] + dp[i][0][j] + dp[i][1][j] + dp[i][2][j]) % mod;
      dp[i + 1][0][j + 1] = (dp[i + 1][0][j + 1] + dp[i][3][j]) % mod;
      dp[i + 1][1][j] = (dp[i + 1][1][j] + dp[i][1][j]) % mod;
      dp[i + 1][1][j + 1] =
          (dp[i + 1][1][j + 1] + dp[i][3][j] + dp[i][0][j]) % mod;
      dp[i + 1][1][j + 2] = (dp[i + 1][1][j + 2] + dp[i][2][j]) % mod;
      dp[i + 1][2][j] = (dp[i + 1][2][j] + dp[i][2][j]) % mod;
      dp[i + 1][2][j + 1] =
          (dp[i + 1][2][j + 1] + dp[i][3][j] + dp[i][0][j]) % mod;
      dp[i + 1][2][j + 2] = (dp[i + 1][2][j + 2] + dp[i][1][j]) % mod;
      dp[i + 1][3][j] =
          (dp[i + 1][3][j] + dp[i][3][j] + dp[i][1][j] + dp[i][2][j]) % mod;
      dp[i + 1][3][j + 1] = (dp[i + 1][3][j + 1] + dp[i][0][j]) % mod;
    }
  }
  cout << (dp[n][0][k] + dp[n][1][k] + dp[n][2][k] + dp[n][3][k]) % mod;
  return 0;
}

### Prompt

Generate 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 mod = 998244353;
long long dp[1001][4][4001];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  dp[1][0][1] = 1;
  dp[1][1][2] = 1;
  dp[1][2][2] = 1;
  dp[1][3][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      dp[i + 1][0][j] =
          (dp[i + 1][0][j] + dp[i][0][j] + dp[i][1][j] + dp[i][2][j]) % mod;
      dp[i + 1][0][j + 1] = (dp[i + 1][0][j + 1] + dp[i][3][j]) % mod;
      dp[i + 1][1][j] = (dp[i + 1][1][j] + dp[i][1][j]) % mod;
      dp[i + 1][1][j + 1] =
          (dp[i + 1][1][j + 1] + dp[i][3][j] + dp[i][0][j]) % mod;
      dp[i + 1][1][j + 2] = (dp[i + 1][1][j + 2] + dp[i][2][j]) % mod;
      dp[i + 1][2][j] = (dp[i + 1][2][j] + dp[i][2][j]) % mod;
      dp[i + 1][2][j + 1] =
          (dp[i + 1][2][j + 1] + dp[i][3][j] + dp[i][0][j]) % mod;
      dp[i + 1][2][j + 2] = (dp[i + 1][2][j + 2] + dp[i][1][j]) % mod;
      dp[i + 1][3][j] =
          (dp[i + 1][3][j] + dp[i][3][j] + dp[i][1][j] + dp[i][2][j]) % mod;
      dp[i + 1][3][j + 1] = (dp[i + 1][3][j + 1] + dp[i][0][j]) % mod;
    }
  }
  cout << (dp[n][0][k] + dp[n][1][k] + dp[n][2][k] + dp[n][3][k]) % mod;
  return 0;
}

```