#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
long long w[3000][2][2], w1[3000][2][2], a, b;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b;
  w[1][1][1] = 1;
  w[1][0][0] = 1;
  w[2][1][0] = 1;
  w[2][0][1] = 1;
  for (int i = 2; i <= a; i++) {
    for (int z = 0; z <= 2000; z++) {
      w1[z][1][1] += w[z][1][1] + w[z][1][0] + w[z][0][1];
      w1[z + 1][1][1] += w[z][0][0];
      w1[z][0][1] += w[z][0][1];
      w1[z + 2][0][1] += w[z][1][0];
      w1[z + 1][0][1] += w[z][1][1];
      w1[z + 1][0][1] += w[z][0][0];
      w1[z][1][0] += w[z][1][0];
      w1[z + 1][1][0] += w[z][1][1] + w[z][0][0];
      w1[z + 2][1][0] += w[z][0][1];
      w1[z][0][0] += w[z][0][0] + w[z][1][0] + w[z][0][1];
      w1[z + 1][0][0] += w[z][1][1];
    }
    for (int z = 0; z <= 2000; z++) {
      w[z][0][0] = w1[z][0][0] % mod;
      w[z][1][0] = w1[z][1][0] % mod;
      w[z][0][1] = w1[z][0][1] % mod;
      w[z][1][1] = w1[z][1][1] % mod;
      w1[z][0][0] = 0;
      w1[z][1][0] = 0;
      w1[z][0][1] = 0;
      w1[z][1][1] = 0;
    }
  }
  cout << (w[b][0][0] + w[b][1][0] + w[b][0][1] + w[b][1][1]) % mod << endl;
  return 0;
}

### 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;
long long mod = 998244353;
long long w[3000][2][2], w1[3000][2][2], a, b;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b;
  w[1][1][1] = 1;
  w[1][0][0] = 1;
  w[2][1][0] = 1;
  w[2][0][1] = 1;
  for (int i = 2; i <= a; i++) {
    for (int z = 0; z <= 2000; z++) {
      w1[z][1][1] += w[z][1][1] + w[z][1][0] + w[z][0][1];
      w1[z + 1][1][1] += w[z][0][0];
      w1[z][0][1] += w[z][0][1];
      w1[z + 2][0][1] += w[z][1][0];
      w1[z + 1][0][1] += w[z][1][1];
      w1[z + 1][0][1] += w[z][0][0];
      w1[z][1][0] += w[z][1][0];
      w1[z + 1][1][0] += w[z][1][1] + w[z][0][0];
      w1[z + 2][1][0] += w[z][0][1];
      w1[z][0][0] += w[z][0][0] + w[z][1][0] + w[z][0][1];
      w1[z + 1][0][0] += w[z][1][1];
    }
    for (int z = 0; z <= 2000; z++) {
      w[z][0][0] = w1[z][0][0] % mod;
      w[z][1][0] = w1[z][1][0] % mod;
      w[z][0][1] = w1[z][0][1] % mod;
      w[z][1][1] = w1[z][1][1] % mod;
      w1[z][0][0] = 0;
      w1[z][1][0] = 0;
      w1[z][0][1] = 0;
      w1[z][1][1] = 0;
    }
  }
  cout << (w[b][0][0] + w[b][1][0] + w[b][0][1] + w[b][1][1]) % mod << endl;
  return 0;
}

```