#include <bits/stdc++.h>
using namespace std;
int tbl[4][4] = {0, 1, 1, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0};
int dp[1500][3000][4];
int main() {
  int n, k;
  cin >> n >> k;
  dp[0][1][0] = 1;
  dp[0][0][0] = 0;
  dp[0][1][3] = 1;
  dp[0][0][1] = 0;
  dp[0][0][2] = 0;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < k + 1; j++) {
      for (int l = 0; l < 4; l++) {
        for (int m = 0; m < 4; m++) {
          int tt = j - tbl[l][m];
          if (tt >= 0) dp[i][j][m] += dp[i - 1][tt][l];
          dp[i][j][m] %= 998244353;
        }
      }
    }
  }
  int pp = 0;
  for (int i = 0; i < 4; i++) {
    pp += dp[n - 1][k][i];
    pp %= 998244353;
  }
  cout << pp;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int tbl[4][4] = {0, 1, 1, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0};
int dp[1500][3000][4];
int main() {
  int n, k;
  cin >> n >> k;
  dp[0][1][0] = 1;
  dp[0][0][0] = 0;
  dp[0][1][3] = 1;
  dp[0][0][1] = 0;
  dp[0][0][2] = 0;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < k + 1; j++) {
      for (int l = 0; l < 4; l++) {
        for (int m = 0; m < 4; m++) {
          int tt = j - tbl[l][m];
          if (tt >= 0) dp[i][j][m] += dp[i - 1][tt][l];
          dp[i][j][m] %= 998244353;
        }
      }
    }
  }
  int pp = 0;
  for (int i = 0; i < 4; i++) {
    pp += dp[n - 1][k][i];
    pp %= 998244353;
  }
  cout << pp;
  return 0;
}

```