#include <bits/stdc++.h>
using namespace std;
long long n, m, f[1100][2100][4];
int main() {
  scanf("%lld%lld", &n, &m);
  memset(f, 0, sizeof(f));
  f[1][2][2] = 1;
  f[1][2][1] = 1;
  f[1][1][0] = 1;
  f[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      for (int k = 0; k <= 3; k++)
        for (int l = 0; l <= 3; l++) {
          int c = 2;
          c -= ((k & 1) == (l & 1));
          c -= ((k & 2) == (l & 2));
          c -= !(((bool)(l & 1)) ^ ((bool)(l & 2)));
          c = (c < 0) ? 0 : c;
          if (j > c) {
            f[i][j][l] = (f[i - 1][j - c][k] + f[i][j][l]) % 998244353;
          }
        }
    }
  }
  long long ans =
      (f[n][m][0] + f[n][m][1] + f[n][m][2] + f[n][m][3]) % 998244353;
  printf("%lld", ans);
}

### Prompt

Develop a solution in CPP 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;
long long n, m, f[1100][2100][4];
int main() {
  scanf("%lld%lld", &n, &m);
  memset(f, 0, sizeof(f));
  f[1][2][2] = 1;
  f[1][2][1] = 1;
  f[1][1][0] = 1;
  f[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      for (int k = 0; k <= 3; k++)
        for (int l = 0; l <= 3; l++) {
          int c = 2;
          c -= ((k & 1) == (l & 1));
          c -= ((k & 2) == (l & 2));
          c -= !(((bool)(l & 1)) ^ ((bool)(l & 2)));
          c = (c < 0) ? 0 : c;
          if (j > c) {
            f[i][j][l] = (f[i - 1][j - c][k] + f[i][j][l]) % 998244353;
          }
        }
    }
  }
  long long ans =
      (f[n][m][0] + f[n][m][1] + f[n][m][2] + f[n][m][3]) % 998244353;
  printf("%lld", ans);
}

```