#include <bits/stdc++.h>
using namespace std;
const int N = (1 << 10);
const long long MOD = 998244353;
long long n, k;
long long dp[N][2 * N][5];
inline void add(long long &aa, long long bb) {
  aa += bb;
  if (aa >= MOD) aa -= MOD;
}
long long solve(long long pos, long long left, int prev) {
  if (pos == n) return (left ? 0 : 1);
  if (dp[pos][left][prev] != -1) return dp[pos][left][prev];
  long long ans = 0;
  if (prev == 4) {
    add(ans, solve(pos + 1, left - 1, 0));
    add(ans, solve(pos + 1, left - 1, 1));
    add(ans, solve(pos + 1, left - 2, 2));
    add(ans, solve(pos + 1, left - 2, 3));
  } else if (prev == 0) {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left - 1, 1));
    add(ans, solve(pos + 1, left - 1, 2));
    add(ans, solve(pos + 1, left - 1, 3));
  } else if (prev == 1) {
    add(ans, solve(pos + 1, left - 1, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left - 1, 2));
    add(ans, solve(pos + 1, left - 1, 3));
  } else if (prev == 2) {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left, 2));
    add(ans, solve(pos + 1, left - 2, 3));
  } else {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left - 2, 2));
    add(ans, solve(pos + 1, left, 3));
  }
  return dp[pos][left][prev] = ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> k;
  memset(dp, (long long)(-1), sizeof dp);
  cout << solve(0, k, 4) << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = (1 << 10);
const long long MOD = 998244353;
long long n, k;
long long dp[N][2 * N][5];
inline void add(long long &aa, long long bb) {
  aa += bb;
  if (aa >= MOD) aa -= MOD;
}
long long solve(long long pos, long long left, int prev) {
  if (pos == n) return (left ? 0 : 1);
  if (dp[pos][left][prev] != -1) return dp[pos][left][prev];
  long long ans = 0;
  if (prev == 4) {
    add(ans, solve(pos + 1, left - 1, 0));
    add(ans, solve(pos + 1, left - 1, 1));
    add(ans, solve(pos + 1, left - 2, 2));
    add(ans, solve(pos + 1, left - 2, 3));
  } else if (prev == 0) {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left - 1, 1));
    add(ans, solve(pos + 1, left - 1, 2));
    add(ans, solve(pos + 1, left - 1, 3));
  } else if (prev == 1) {
    add(ans, solve(pos + 1, left - 1, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left - 1, 2));
    add(ans, solve(pos + 1, left - 1, 3));
  } else if (prev == 2) {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left, 2));
    add(ans, solve(pos + 1, left - 2, 3));
  } else {
    add(ans, solve(pos + 1, left, 0));
    add(ans, solve(pos + 1, left, 1));
    add(ans, solve(pos + 1, left - 2, 2));
    add(ans, solve(pos + 1, left, 3));
  }
  return dp[pos][left][prev] = ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> k;
  memset(dp, (long long)(-1), sizeof dp);
  cout << solve(0, k, 4) << '\n';
  return 0;
}

```