#include <bits/stdc++.h>
using namespace std;
const unsigned int mod = 998244353;
unsigned int ww[1001][2001], bw[1001][2001];
unsigned int dpbw(int n, int k);
unsigned int dpww(int n, int k) {
  if (k < 1 or k > 2 * n) return 0;
  unsigned int &add = ww[n][k];
  if (add) return add;
  return add = (dpww(n - 1, k) + 2 * dpbw(n - 1, k) + dpww(n - 1, k - 1)) % mod;
}
unsigned int dpbw(int n, int k) {
  if (k < 1 or k > 2 * n) return 0;
  unsigned int &add = bw[n][k];
  if (add) return add;
  return add = ((2 * dpww(n - 1, k - 1)) % mod + dpbw(n - 1, k) +
                dpbw(n - 1, k - 2)) %
               mod;
}
int main() {
  ww[1][1] = 2;
  bw[1][2] = 2;
  ww[1][2] = 0;
  bw[1][1] = 0;
  int n, k;
  cin >> n >> k;
  cout << (dpww(n, k) + dpbw(n, k)) % mod << endl;
}

### Prompt

Your challenge is to write 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;
const unsigned int mod = 998244353;
unsigned int ww[1001][2001], bw[1001][2001];
unsigned int dpbw(int n, int k);
unsigned int dpww(int n, int k) {
  if (k < 1 or k > 2 * n) return 0;
  unsigned int &add = ww[n][k];
  if (add) return add;
  return add = (dpww(n - 1, k) + 2 * dpbw(n - 1, k) + dpww(n - 1, k - 1)) % mod;
}
unsigned int dpbw(int n, int k) {
  if (k < 1 or k > 2 * n) return 0;
  unsigned int &add = bw[n][k];
  if (add) return add;
  return add = ((2 * dpww(n - 1, k - 1)) % mod + dpbw(n - 1, k) +
                dpbw(n - 1, k - 2)) %
               mod;
}
int main() {
  ww[1][1] = 2;
  bw[1][2] = 2;
  ww[1][2] = 0;
  bw[1][1] = 0;
  int n, k;
  cin >> n >> k;
  cout << (dpww(n, k) + dpbw(n, k)) % mod << endl;
}

```