#include <bits/stdc++.h>
using namespace std;
int n, comp;
int mat[4][4] = {{0, 1, 1, 1}, {0, 0, 2, 0}, {0, 2, 0, 0}, {1, 1, 1, 0}};
long long dp[1010][2010][4];
int solve(int pos, int K, int prev) {
  if (pos == n) return K == 0;
  if (dp[pos][K][prev] != -1) return dp[pos][K][prev];
  int ans = 0;
  for (int i = 0; i < 4; ++i)
    ans = (ans + solve(pos + 1, K - mat[prev][i], i)) % 998244353LL;
  return (dp[pos][K][prev] = ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> comp;
  memset(dp, -1, sizeof(dp));
  long long ans = 0;
  for (int i = 0; i < 4; ++i)
    ans = (ans + solve(1, comp - 1 - (i == 1 || i == 2), i)) % 998244353LL;
  cout << ans << '\n';
  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;
int n, comp;
int mat[4][4] = {{0, 1, 1, 1}, {0, 0, 2, 0}, {0, 2, 0, 0}, {1, 1, 1, 0}};
long long dp[1010][2010][4];
int solve(int pos, int K, int prev) {
  if (pos == n) return K == 0;
  if (dp[pos][K][prev] != -1) return dp[pos][K][prev];
  int ans = 0;
  for (int i = 0; i < 4; ++i)
    ans = (ans + solve(pos + 1, K - mat[prev][i], i)) % 998244353LL;
  return (dp[pos][K][prev] = ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> comp;
  memset(dp, -1, sizeof(dp));
  long long ans = 0;
  for (int i = 0; i < 4; ++i)
    ans = (ans + solve(1, comp - 1 - (i == 1 || i == 2), i)) % 998244353LL;
  cout << ans << '\n';
  return 0;
}

```