#include <bits/stdc++.h>
using namespace std;
const int N = 1005, mod = 998244353;
int n, k, dp[2 * N][N][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    if (i == 1) {
      for (int j = 1; j <= n; j++) dp[i][j][0] = 2;
      continue;
    }
    if (i == 2) dp[i][1][1] = 2;
    for (int j = 2; j <= n; j++) {
      dp[i][j][0] = ((long long)dp[i][j - 1][0] + 2 * dp[i][j - 1][1] +
                     dp[i - 1][j - 1][0]) %
                    mod;
      dp[i][j][1] = ((long long)dp[i][j - 1][1] + 2 * dp[i - 1][j - 1][0] +
                     dp[i - 2][j - 1][1]) %
                    mod;
    }
  }
  dp[k][n][0] = (dp[k][n][0] + dp[k][n][1]) % mod;
  cout << dp[k][n][0] << endl;
  return 0;
}

### Prompt

Please formulate 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 N = 1005, mod = 998244353;
int n, k, dp[2 * N][N][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    if (i == 1) {
      for (int j = 1; j <= n; j++) dp[i][j][0] = 2;
      continue;
    }
    if (i == 2) dp[i][1][1] = 2;
    for (int j = 2; j <= n; j++) {
      dp[i][j][0] = ((long long)dp[i][j - 1][0] + 2 * dp[i][j - 1][1] +
                     dp[i - 1][j - 1][0]) %
                    mod;
      dp[i][j][1] = ((long long)dp[i][j - 1][1] + 2 * dp[i - 1][j - 1][0] +
                     dp[i - 2][j - 1][1]) %
                    mod;
    }
  }
  dp[k][n][0] = (dp[k][n][0] + dp[k][n][1]) % mod;
  cout << dp[k][n][0] << endl;
  return 0;
}

```