#include <bits/stdc++.h>
using namespace std;
const int maxn = 1100;
long long dp[maxn][2 * maxn][2], n, k;
int main() {
  dp[1][1][0] = 2;
  dp[1][2][1] = 2;
  for (long long i = 1; i < maxn - 3; i++) {
    for (long long j = 1; j <= 2 * i; j++) {
      dp[i + 1][j][0] += dp[i][j][0];
      dp[i + 1][j + 1][0] += dp[i][j][0];
      dp[i + 1][j + 1][1] += (2 * dp[i][j][0]) % 998244353;
      dp[i + 1][j][0] += (2 * dp[i][j][1]) % 998244353;
      dp[i + 1][j][1] += dp[i][j][1];
      dp[i + 1][j + 2][1] += dp[i][j][1];
      dp[i + 1][j + 2][1] %= 998244353;
      dp[i + 1][j][1] %= 998244353;
      dp[i + 1][j][0] %= 998244353;
      dp[i + 1][j + 1][1] %= 998244353;
      dp[i + 1][j + 1][0] %= 998244353;
      dp[i + 1][j][0] %= 998244353;
    }
  }
  cin >> n >> k;
  cout << ((dp[n][k][0]) + (dp[n][k][1]) + 998244353) % 998244353;
}

### Prompt

Your task is to create 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;
const int maxn = 1100;
long long dp[maxn][2 * maxn][2], n, k;
int main() {
  dp[1][1][0] = 2;
  dp[1][2][1] = 2;
  for (long long i = 1; i < maxn - 3; i++) {
    for (long long j = 1; j <= 2 * i; j++) {
      dp[i + 1][j][0] += dp[i][j][0];
      dp[i + 1][j + 1][0] += dp[i][j][0];
      dp[i + 1][j + 1][1] += (2 * dp[i][j][0]) % 998244353;
      dp[i + 1][j][0] += (2 * dp[i][j][1]) % 998244353;
      dp[i + 1][j][1] += dp[i][j][1];
      dp[i + 1][j + 2][1] += dp[i][j][1];
      dp[i + 1][j + 2][1] %= 998244353;
      dp[i + 1][j][1] %= 998244353;
      dp[i + 1][j][0] %= 998244353;
      dp[i + 1][j + 1][1] %= 998244353;
      dp[i + 1][j + 1][0] %= 998244353;
      dp[i + 1][j][0] %= 998244353;
    }
  }
  cin >> n >> k;
  cout << ((dp[n][k][0]) + (dp[n][k][1]) + 998244353) % 998244353;
}

```