#include <bits/stdc++.h>
using namespace std;
long long dp[1000][2000][4];
int f[4][4] = {0, 1, 1, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0};
int main() {
  dp[0][1][0] = 1;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  dp[0][1][3] = 1;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j <= k; j++) {
      for (int s = 0; s < 4; s++) {
        for (int s2 = 0; s2 < 4; s2++) {
          dp[i + 1][j + f[s][s2]][s2] =
              (dp[i + 1][j + f[s][s2]][s2] + dp[i][j][s]) % 998244353;
        }
      }
    }
  }
  cout << (dp[n - 1][k][0] + dp[n - 1][k][1] + dp[n - 1][k][2] +
           dp[n - 1][k][3]) %
              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>
using namespace std;
long long dp[1000][2000][4];
int f[4][4] = {0, 1, 1, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0};
int main() {
  dp[0][1][0] = 1;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  dp[0][1][3] = 1;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j <= k; j++) {
      for (int s = 0; s < 4; s++) {
        for (int s2 = 0; s2 < 4; s2++) {
          dp[i + 1][j + f[s][s2]][s2] =
              (dp[i + 1][j + f[s][s2]][s2] + dp[i][j][s]) % 998244353;
        }
      }
    }
  }
  cout << (dp[n - 1][k][0] + dp[n - 1][k][1] + dp[n - 1][k][2] +
           dp[n - 1][k][3]) %
              998244353;
}

```