#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const int K = 2005;
using i64 = int64_t;
const i64 MOD = 998244353;
i64 f[K][N][4];
int main() {
  int n, K;
  scanf("%d%d", &n, &K);
  f[1][0][0] = f[1][0][3] = 1;
  f[2][0][1] = f[2][0][2] = 1;
  for (int k = 0; k <= K; ++k) {
    for (int i = 1; i < n; ++i) {
      f[k][i][0] = ((k - 0 >= 0 ? f[k - 0][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][1] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 2 >= 0 ? f[k - 2][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][2] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 2 >= 0 ? f[k - 2][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][3] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][3] : 0)) %
                   MOD;
    }
  }
  --n;
  i64 ans = (f[K][n][0] + f[K][n][1] + f[K][n][2] + f[K][n][3]) % MOD;
  printf("%d\n", (int)ans);
  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>
using namespace std;
const int N = 1005;
const int K = 2005;
using i64 = int64_t;
const i64 MOD = 998244353;
i64 f[K][N][4];
int main() {
  int n, K;
  scanf("%d%d", &n, &K);
  f[1][0][0] = f[1][0][3] = 1;
  f[2][0][1] = f[2][0][2] = 1;
  for (int k = 0; k <= K; ++k) {
    for (int i = 1; i < n; ++i) {
      f[k][i][0] = ((k - 0 >= 0 ? f[k - 0][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][1] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 2 >= 0 ? f[k - 2][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][2] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 2 >= 0 ? f[k - 2][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 1 >= 0 ? f[k - 1][i - 1][3] : 0)) %
                   MOD;
      f[k][i][3] = ((k - 1 >= 0 ? f[k - 1][i - 1][0] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][1] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][2] : 0) +
                    (k - 0 >= 0 ? f[k - 0][i - 1][3] : 0)) %
                   MOD;
    }
  }
  --n;
  i64 ans = (f[K][n][0] + f[K][n][1] + f[K][n][2] + f[K][n][3]) % MOD;
  printf("%d\n", (int)ans);
  return 0;
}

```