#include <bits/stdc++.h>
long long dp[1005][4][2005];
int main(void) {
  int n, j, i, k;
  scanf("%d%d", &n, &k);
  dp[1][0][1] = dp[1][3][1] = 1;
  dp[1][1][2] = dp[1][2][2] = 1;
  for (i = 2; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      dp[i][0][j] = (dp[i - 1][0][j] + dp[i - 1][1][j] + dp[i - 1][2][j] +
                     dp[i - 1][3][j - 1]) %
                    998244353;
      dp[i][3][j] = (dp[i - 1][1][j] + dp[i - 1][2][j] + dp[i - 1][3][j] +
                     dp[i - 1][0][j - 1]) %
                    998244353;
      dp[i][1][j] =
          (dp[i - 1][0][j - 1] + dp[i - 1][1][j] + dp[i - 1][3][j - 1]) %
          998244353;
      dp[i][2][j] =
          (dp[i - 1][0][j - 1] + dp[i - 1][2][j] + dp[i - 1][3][j - 1]) %
          998244353;
      if (j >= 2) {
        dp[i][1][j] = (dp[i][1][j] + dp[i - 1][2][j - 2]) % 998244353;
        dp[i][2][j] = (dp[i][2][j] + dp[i - 1][1][j - 2]) % 998244353;
      }
    }
  }
  printf("%lld\n",
         (dp[n][0][k] + dp[n][1][k] + dp[n][2][k] + dp[n][3][k]) % 998244353);
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
long long dp[1005][4][2005];
int main(void) {
  int n, j, i, k;
  scanf("%d%d", &n, &k);
  dp[1][0][1] = dp[1][3][1] = 1;
  dp[1][1][2] = dp[1][2][2] = 1;
  for (i = 2; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      dp[i][0][j] = (dp[i - 1][0][j] + dp[i - 1][1][j] + dp[i - 1][2][j] +
                     dp[i - 1][3][j - 1]) %
                    998244353;
      dp[i][3][j] = (dp[i - 1][1][j] + dp[i - 1][2][j] + dp[i - 1][3][j] +
                     dp[i - 1][0][j - 1]) %
                    998244353;
      dp[i][1][j] =
          (dp[i - 1][0][j - 1] + dp[i - 1][1][j] + dp[i - 1][3][j - 1]) %
          998244353;
      dp[i][2][j] =
          (dp[i - 1][0][j - 1] + dp[i - 1][2][j] + dp[i - 1][3][j - 1]) %
          998244353;
      if (j >= 2) {
        dp[i][1][j] = (dp[i][1][j] + dp[i - 1][2][j - 2]) % 998244353;
        dp[i][2][j] = (dp[i][2][j] + dp[i - 1][1][j - 2]) % 998244353;
      }
    }
  }
  printf("%lld\n",
         (dp[n][0][k] + dp[n][1][k] + dp[n][2][k] + dp[n][3][k]) % 998244353);
  return 0;
}

```