#include <bits/stdc++.h>
using namespace std;
ifstream fin("AAtest.in.txt");
long long n, k, jad[4][2005], jad1[4][2005], mit;
long long c = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n >> k;
  jad[0][1] = 1;
  jad[1][2] = 1;
  jad[2][2] = 1;
  jad[3][1] = 1;
  long long sum = 0;
  for (int(i) = (0); ((i)) < ((n - 1)); ((i))++) {
    for (int j = 1; j <= k; j++) {
      jad1[0][j] = jad[0][j] + jad[1][j] + jad[2][j] + jad[3][j - 1];
      jad1[0][j] %= c;
      jad1[1][j] = jad[1][j] + jad[0][j - 1] + jad[3][j - 1];
      if (j > 3) jad1[1][j] += jad[2][j - 2];
      jad1[1][j] %= c;
      jad1[2][j] = jad[2][j] + jad[0][j - 1] + jad[3][j - 1];
      if (j > 3) jad1[2][j] += jad[1][j - 2];
      jad1[2][j] %= c;
      jad1[3][j] = jad[3][j] + jad[0][j - 1] + jad[1][j] + jad[2][j];
      jad1[3][j] %= c;
    }
    for (int(i) = (0); ((i)) < ((k + 1)); ((i))++)
      for (int(j) = (0); ((j)) < ((4)); ((j))++) jad[j][i] = jad1[j][i];
    for (int(i) = (0); ((i)) < ((4)); ((i))++) memset(jad1[i], 0, k + 1);
  }
  for (int(i) = (0); ((i)) < ((4)); ((i))++) {
    sum += jad[i][k];
    sum %= c;
  }
  cout << sum << endl;
}

### Prompt

Your task is to create 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;
ifstream fin("AAtest.in.txt");
long long n, k, jad[4][2005], jad1[4][2005], mit;
long long c = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n >> k;
  jad[0][1] = 1;
  jad[1][2] = 1;
  jad[2][2] = 1;
  jad[3][1] = 1;
  long long sum = 0;
  for (int(i) = (0); ((i)) < ((n - 1)); ((i))++) {
    for (int j = 1; j <= k; j++) {
      jad1[0][j] = jad[0][j] + jad[1][j] + jad[2][j] + jad[3][j - 1];
      jad1[0][j] %= c;
      jad1[1][j] = jad[1][j] + jad[0][j - 1] + jad[3][j - 1];
      if (j > 3) jad1[1][j] += jad[2][j - 2];
      jad1[1][j] %= c;
      jad1[2][j] = jad[2][j] + jad[0][j - 1] + jad[3][j - 1];
      if (j > 3) jad1[2][j] += jad[1][j - 2];
      jad1[2][j] %= c;
      jad1[3][j] = jad[3][j] + jad[0][j - 1] + jad[1][j] + jad[2][j];
      jad1[3][j] %= c;
    }
    for (int(i) = (0); ((i)) < ((k + 1)); ((i))++)
      for (int(j) = (0); ((j)) < ((4)); ((j))++) jad[j][i] = jad1[j][i];
    for (int(i) = (0); ((i)) < ((4)); ((i))++) memset(jad1[i], 0, k + 1);
  }
  for (int(i) = (0); ((i)) < ((4)); ((i))++) {
    sum += jad[i][k];
    sum %= c;
  }
  cout << sum << endl;
}

```