#include <bits/stdc++.h>
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long a[2008] = {0, 2};
  long long b[2008] = {0, 0, 2};
  while (--n)
    for (int i = k; i >= 1; --i) {
      a[i] = (a[i] + a[i - 1] + 2 * b[i]) % 998244353;
      b[i] = (b[i] + b[i - 2] + 2 * a[i - 1]) % 998244353;
    }
  printf("%d", (a[k] + b[k]) % 998244353);
}

### 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>
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long a[2008] = {0, 2};
  long long b[2008] = {0, 0, 2};
  while (--n)
    for (int i = k; i >= 1; --i) {
      a[i] = (a[i] + a[i - 1] + 2 * b[i]) % 998244353;
      b[i] = (b[i] + b[i - 2] + 2 * a[i - 1]) % 998244353;
    }
  printf("%d", (a[k] + b[k]) % 998244353);
}

```