#include <bits/stdc++.h>
using namespace std;
long long dp[1010][4010][4];
int main() {
  int j, p;
  cin >> j >> p;
  dp[1][1][0] = 1;
  dp[1][1][1] = 1;
  dp[1][2][2] = 1;
  dp[1][2][3] = 1;
  for (int n = 1; n <= j; n++) {
    for (int k = 1; k <= p; k++) {
      dp[n + 1][k][0] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][0] += dp[n][k][1] % 998244353;
      dp[n + 1][k][0] += dp[n][k][2] % 998244353;
      dp[n + 1][k][0] += dp[n][k][3] % 998244353;
      dp[n + 1][k][1] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 1][1] += dp[n][k][0] % 998244353;
      dp[n + 1][k][1] += dp[n][k][2] % 998244353;
      dp[n + 1][k][1] += dp[n][k][3] % 998244353;
      dp[n + 1][k][2] += dp[n][k][2] % 998244353;
      dp[n + 1][k + 1][2] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][2] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 2][2] += dp[n][k][3] % 998244353;
      dp[n + 1][k][3] += dp[n][k][3] % 998244353;
      dp[n + 1][k + 1][3] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][3] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 2][3] += dp[n][k][2] % 998244353;
    }
  }
  std::cout << (dp[j][p][0] % 998244353 + dp[j][p][1] % 998244353 +
                dp[j][p][2] % 998244353 + dp[j][p][3] % 998244353) %
                   998244353
            << endl;
}

### Prompt

Develop a solution in CPP 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;
long long dp[1010][4010][4];
int main() {
  int j, p;
  cin >> j >> p;
  dp[1][1][0] = 1;
  dp[1][1][1] = 1;
  dp[1][2][2] = 1;
  dp[1][2][3] = 1;
  for (int n = 1; n <= j; n++) {
    for (int k = 1; k <= p; k++) {
      dp[n + 1][k][0] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][0] += dp[n][k][1] % 998244353;
      dp[n + 1][k][0] += dp[n][k][2] % 998244353;
      dp[n + 1][k][0] += dp[n][k][3] % 998244353;
      dp[n + 1][k][1] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 1][1] += dp[n][k][0] % 998244353;
      dp[n + 1][k][1] += dp[n][k][2] % 998244353;
      dp[n + 1][k][1] += dp[n][k][3] % 998244353;
      dp[n + 1][k][2] += dp[n][k][2] % 998244353;
      dp[n + 1][k + 1][2] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][2] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 2][2] += dp[n][k][3] % 998244353;
      dp[n + 1][k][3] += dp[n][k][3] % 998244353;
      dp[n + 1][k + 1][3] += dp[n][k][0] % 998244353;
      dp[n + 1][k + 1][3] += dp[n][k][1] % 998244353;
      dp[n + 1][k + 2][3] += dp[n][k][2] % 998244353;
    }
  }
  std::cout << (dp[j][p][0] % 998244353 + dp[j][p][1] % 998244353 +
                dp[j][p][2] % 998244353 + dp[j][p][3] % 998244353) %
                   998244353
            << endl;
}

```