#include <bits/stdc++.h>
using namespace std;
long long n, k, dp[1002][2002][4][4], mod = 998244353;
long long solve(long long i, long long fo2, long long t7t, long long rem) {
  if (rem < 0) return 0;
  if (dp[i][rem][fo2][t7t] != -1) return dp[i][rem][fo2][t7t];
  if (i > n) return rem == 0;
  long long ans = 0;
  ans += solve(i + 1, 1, 1, rem - (fo2 != 1 && t7t != 1));
  ans %= mod;
  ans += solve(i + 1, 2, 2, rem - (fo2 != 2 && t7t != 2));
  ans %= mod;
  ans += solve(i + 1, 1, 2, rem - (fo2 != 1) - (t7t != 2));
  ans %= mod;
  ans += solve(i + 1, 2, 1, rem - (fo2 != 2) - (t7t != 1));
  ans %= mod;
  return dp[i][rem][fo2][t7t] = ans;
}
int main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> k;
  cout << solve(1, 0, 0, k) << "\n";
  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;
long long n, k, dp[1002][2002][4][4], mod = 998244353;
long long solve(long long i, long long fo2, long long t7t, long long rem) {
  if (rem < 0) return 0;
  if (dp[i][rem][fo2][t7t] != -1) return dp[i][rem][fo2][t7t];
  if (i > n) return rem == 0;
  long long ans = 0;
  ans += solve(i + 1, 1, 1, rem - (fo2 != 1 && t7t != 1));
  ans %= mod;
  ans += solve(i + 1, 2, 2, rem - (fo2 != 2 && t7t != 2));
  ans %= mod;
  ans += solve(i + 1, 1, 2, rem - (fo2 != 1) - (t7t != 2));
  ans %= mod;
  ans += solve(i + 1, 2, 1, rem - (fo2 != 2) - (t7t != 1));
  ans %= mod;
  return dp[i][rem][fo2][t7t] = ans;
}
int main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> k;
  cout << solve(1, 0, 0, k) << "\n";
  return 0;
}

```