#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int dp[n + 1][k + 1][2][2];
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= k; j++)
      for (int p = 0; p <= 1; p++)
        for (int q = 0; q <= 1; q++) dp[i][j][p][q] = 0;
  for (int p = 0; p <= 1; p++)
    for (int q = 0; q <= 1; q++) dp[0][0][p][q] = 1;
  dp[1][1][0][0] = 1;
  dp[1][1][1][1] = 1;
  dp[1][1][0][1] = 0;
  dp[1][1][1][0] = 0;
  dp[1][2][0][0] = 0;
  dp[1][2][1][1] = 0;
  dp[1][2][0][1] = 1;
  dp[1][2][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][1][0][0] = 1;
    dp[i][1][1][1] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= min(2 * i, k); j++) {
      dp[i][j][0][1] =
          ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][0][0]) % mod +
           (dp[i - 1][j - 1][1][1] + dp[i - 1][j - 2][1][0]) % mod) %
          mod;
      dp[i][j][1][0] =
          ((dp[i - 1][j - 2][0][1] + dp[i - 1][j - 1][0][0]) % mod +
           (dp[i - 1][j][1][0] + dp[i - 1][j - 1][1][1]) % mod) %
          mod;
      dp[i][j][0][0] = ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][1][1]) % mod +
                        (dp[i - 1][j][1][0] + dp[i - 1][j][0][0]) % mod) %
                       mod;
      dp[i][j][1][1] = ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][0][0]) % mod +
                        (dp[i - 1][j][1][0] + dp[i - 1][j][1][1]) % mod) %
                       mod;
    }
  }
  cout << ((dp[n][k][0][0] + dp[n][k][0][1]) % mod +
           (dp[n][k][1][0] + dp[n][k][1][1]) % mod) %
              mod
       << "\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;
const int mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int dp[n + 1][k + 1][2][2];
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= k; j++)
      for (int p = 0; p <= 1; p++)
        for (int q = 0; q <= 1; q++) dp[i][j][p][q] = 0;
  for (int p = 0; p <= 1; p++)
    for (int q = 0; q <= 1; q++) dp[0][0][p][q] = 1;
  dp[1][1][0][0] = 1;
  dp[1][1][1][1] = 1;
  dp[1][1][0][1] = 0;
  dp[1][1][1][0] = 0;
  dp[1][2][0][0] = 0;
  dp[1][2][1][1] = 0;
  dp[1][2][0][1] = 1;
  dp[1][2][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][1][0][0] = 1;
    dp[i][1][1][1] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= min(2 * i, k); j++) {
      dp[i][j][0][1] =
          ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][0][0]) % mod +
           (dp[i - 1][j - 1][1][1] + dp[i - 1][j - 2][1][0]) % mod) %
          mod;
      dp[i][j][1][0] =
          ((dp[i - 1][j - 2][0][1] + dp[i - 1][j - 1][0][0]) % mod +
           (dp[i - 1][j][1][0] + dp[i - 1][j - 1][1][1]) % mod) %
          mod;
      dp[i][j][0][0] = ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][1][1]) % mod +
                        (dp[i - 1][j][1][0] + dp[i - 1][j][0][0]) % mod) %
                       mod;
      dp[i][j][1][1] = ((dp[i - 1][j][0][1] + dp[i - 1][j - 1][0][0]) % mod +
                        (dp[i - 1][j][1][0] + dp[i - 1][j][1][1]) % mod) %
                       mod;
    }
  }
  cout << ((dp[n][k][0][0] + dp[n][k][0][1]) % mod +
           (dp[n][k][1][0] + dp[n][k][1][1]) % mod) %
              mod
       << "\n";
  return 0;
}

```