#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  vector<vector<long long> > dp[4];
  for (int i = 0; i < 4; i++) {
    dp[i].resize(n + 1, vector<long long>(k + 1, 0));
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= k; j++) {
      for (long long p = 0; p < 4; p++) {
        if (i * 2 < j || j == 0) {
          dp[p][i][j] = 0;
        } else if (i == 1) {
          if ((p == 0 || p == 3) && j == 1)
            dp[p][i][j] = 1;
          else if ((p == 1 || p == 2) && j == 2)
            dp[p][i][j] = 1;
          else
            dp[p][i][j] = 0;
        } else {
          if (p == 0) {
            dp[p][i][j] = (dp[0][i - 1][j] + dp[1][i - 1][j] + dp[2][i - 1][j] +
                           dp[3][i - 1][j - 1]) %
                          mod;
          } else if (p == 1) {
            dp[p][i][j] =
                (dp[0][i - 1][j - 1] + dp[1][i - 1][j] + dp[3][i - 1][j - 1]) %
                mod;
            if (j >= 2) {
              dp[p][i][j] = (dp[p][i][j] + dp[2][i - 1][j - 2]) % mod;
            }
          } else if (p == 2) {
            dp[p][i][j] =
                (dp[0][i - 1][j - 1] + dp[2][i - 1][j] + dp[3][i - 1][j - 1]) %
                mod;
            if (j >= 2) {
              dp[p][i][j] = (dp[p][i][j] + dp[1][i - 1][j - 2]) % mod;
            }
          } else if (p == 3) {
            dp[p][i][j] = (dp[0][i - 1][j - 1] + dp[1][i - 1][j] +
                           dp[2][i - 1][j] + dp[3][i - 1][j]) %
                          mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (long long p = 0; p < 4; p++) {
    ans += dp[p][n][k];
    ans %= mod;
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  vector<vector<long long> > dp[4];
  for (int i = 0; i < 4; i++) {
    dp[i].resize(n + 1, vector<long long>(k + 1, 0));
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= k; j++) {
      for (long long p = 0; p < 4; p++) {
        if (i * 2 < j || j == 0) {
          dp[p][i][j] = 0;
        } else if (i == 1) {
          if ((p == 0 || p == 3) && j == 1)
            dp[p][i][j] = 1;
          else if ((p == 1 || p == 2) && j == 2)
            dp[p][i][j] = 1;
          else
            dp[p][i][j] = 0;
        } else {
          if (p == 0) {
            dp[p][i][j] = (dp[0][i - 1][j] + dp[1][i - 1][j] + dp[2][i - 1][j] +
                           dp[3][i - 1][j - 1]) %
                          mod;
          } else if (p == 1) {
            dp[p][i][j] =
                (dp[0][i - 1][j - 1] + dp[1][i - 1][j] + dp[3][i - 1][j - 1]) %
                mod;
            if (j >= 2) {
              dp[p][i][j] = (dp[p][i][j] + dp[2][i - 1][j - 2]) % mod;
            }
          } else if (p == 2) {
            dp[p][i][j] =
                (dp[0][i - 1][j - 1] + dp[2][i - 1][j] + dp[3][i - 1][j - 1]) %
                mod;
            if (j >= 2) {
              dp[p][i][j] = (dp[p][i][j] + dp[1][i - 1][j - 2]) % mod;
            }
          } else if (p == 3) {
            dp[p][i][j] = (dp[0][i - 1][j - 1] + dp[1][i - 1][j] +
                           dp[2][i - 1][j] + dp[3][i - 1][j]) %
                          mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (long long p = 0; p < 4; p++) {
    ans += dp[p][n][k];
    ans %= mod;
  }
  cout << ans;
  return 0;
}

```