#include <bits/stdc++.h>
using namespace std;
long long dp[1005][2 * 1005][2][2];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    if (i == 1) {
      dp[1][2][0][1] = 1;
      dp[1][2][1][0] = 1;
      dp[1][1][0][0] = 1;
      dp[1][1][1][1] = 1;
      continue;
    }
    for (int j = 1; j <= 2 * i; j++) {
      dp[i][j][0][0] += (2 * dp[i - 1][j][0][1]) % 998244353;
      dp[i][j][0][0] %= 998244353;
      dp[i][j][0][0] += dp[i - 1][j][0][0];
      dp[i][j][0][0] %= 998244353;
      dp[i][j][0][0] += dp[i - 1][j - 1][1][1];
      dp[i][j][0][0] %= 998244353;
      dp[i][j][1][1] = dp[i][j][0][0];
      dp[i][j][0][1] += dp[i - 1][j][0][1];
      dp[i][j][0][1] %= 998244353;
      if (j > 1) dp[i][j][0][1] += dp[i - 1][j - 2][1][0];
      dp[i][j][0][1] %= 998244353;
      dp[i][j][0][1] += (2 * dp[i - 1][j - 1][0][0] % 998244353);
      dp[i][j][0][1] %= 998244353;
      dp[i][j][1][0] = dp[i][j][0][1];
    }
  }
  int ans =
      (dp[n][k][0][0] + dp[n][k][0][1] + dp[n][k][1][1] + dp[n][k][1][0]) %
      998244353;
  printf("%d\n", ans);
  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;
long long dp[1005][2 * 1005][2][2];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    if (i == 1) {
      dp[1][2][0][1] = 1;
      dp[1][2][1][0] = 1;
      dp[1][1][0][0] = 1;
      dp[1][1][1][1] = 1;
      continue;
    }
    for (int j = 1; j <= 2 * i; j++) {
      dp[i][j][0][0] += (2 * dp[i - 1][j][0][1]) % 998244353;
      dp[i][j][0][0] %= 998244353;
      dp[i][j][0][0] += dp[i - 1][j][0][0];
      dp[i][j][0][0] %= 998244353;
      dp[i][j][0][0] += dp[i - 1][j - 1][1][1];
      dp[i][j][0][0] %= 998244353;
      dp[i][j][1][1] = dp[i][j][0][0];
      dp[i][j][0][1] += dp[i - 1][j][0][1];
      dp[i][j][0][1] %= 998244353;
      if (j > 1) dp[i][j][0][1] += dp[i - 1][j - 2][1][0];
      dp[i][j][0][1] %= 998244353;
      dp[i][j][0][1] += (2 * dp[i - 1][j - 1][0][0] % 998244353);
      dp[i][j][0][1] %= 998244353;
      dp[i][j][1][0] = dp[i][j][0][1];
    }
  }
  int ans =
      (dp[n][k][0][0] + dp[n][k][0][1] + dp[n][k][1][1] + dp[n][k][1][0]) %
      998244353;
  printf("%d\n", ans);
  return 0;
}

```