#include <bits/stdc++.h>
using namespace std;
const int N = 1100;
const int MAXN = int(1e6) + 100;
const long long mod = 998244353;
int solution();
int main(int argc, char* const argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return solution();
}
int n, k;
long long dp[4][1001][2001];
int solution() {
  cin >> n >> k;
  dp[0][1][1] = 1;
  dp[1][1][2] = 1;
  dp[2][1][2] = 1;
  dp[3][1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      (dp[0][i + 1][j] += dp[0][i][j]) %= mod;
      (dp[0][i + 1][j] += dp[1][i][j]) %= mod;
      (dp[0][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[0][i + 1][j + 1] += dp[3][i][j]) %= mod;
      (dp[1][i + 1][j] += dp[1][i][j]) %= mod;
      if (j + 1 <= k) (dp[1][i + 1][j + 1] += dp[0][i][j]) %= mod;
      if (j + 1 <= k) (dp[1][i + 1][j + 1] += dp[3][i][j]) %= mod;
      if (j + 2 <= k) (dp[1][i + 1][j + 2] += dp[2][i][j]) %= mod;
      (dp[2][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[2][i + 1][j + 1] += dp[0][i][j]) %= mod;
      if (j + 1 <= k) (dp[2][i + 1][j + 1] += dp[3][i][j]) %= mod;
      if (j + 2 <= k) (dp[2][i + 1][j + 2] += dp[1][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[3][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[1][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[3][i + 1][j + 1] += dp[0][i][j]) %= mod;
    }
  }
  (dp[0][n][k] += dp[1][n][k]) %= mod;
  (dp[0][n][k] += dp[2][n][k]) %= mod;
  (dp[0][n][k] += dp[3][n][k]) %= mod;
  cout << dp[0][n][k];
  return 0;
}

### 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 N = 1100;
const int MAXN = int(1e6) + 100;
const long long mod = 998244353;
int solution();
int main(int argc, char* const argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return solution();
}
int n, k;
long long dp[4][1001][2001];
int solution() {
  cin >> n >> k;
  dp[0][1][1] = 1;
  dp[1][1][2] = 1;
  dp[2][1][2] = 1;
  dp[3][1][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      (dp[0][i + 1][j] += dp[0][i][j]) %= mod;
      (dp[0][i + 1][j] += dp[1][i][j]) %= mod;
      (dp[0][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[0][i + 1][j + 1] += dp[3][i][j]) %= mod;
      (dp[1][i + 1][j] += dp[1][i][j]) %= mod;
      if (j + 1 <= k) (dp[1][i + 1][j + 1] += dp[0][i][j]) %= mod;
      if (j + 1 <= k) (dp[1][i + 1][j + 1] += dp[3][i][j]) %= mod;
      if (j + 2 <= k) (dp[1][i + 1][j + 2] += dp[2][i][j]) %= mod;
      (dp[2][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[2][i + 1][j + 1] += dp[0][i][j]) %= mod;
      if (j + 1 <= k) (dp[2][i + 1][j + 1] += dp[3][i][j]) %= mod;
      if (j + 2 <= k) (dp[2][i + 1][j + 2] += dp[1][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[3][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[1][i][j]) %= mod;
      (dp[3][i + 1][j] += dp[2][i][j]) %= mod;
      if (j + 1 <= k) (dp[3][i + 1][j + 1] += dp[0][i][j]) %= mod;
    }
  }
  (dp[0][n][k] += dp[1][n][k]) %= mod;
  (dp[0][n][k] += dp[2][n][k]) %= mod;
  (dp[0][n][k] += dp[3][n][k]) %= mod;
  cout << dp[0][n][k];
  return 0;
}

```