#include <bits/stdc++.h>
const int mod = 998244353;
long long dp[1005][3][3][2005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[1][1][1][0] = 0;
  dp[1][1][2][0] = 0;
  dp[1][2][1][0] = 0;
  dp[1][2][2][0] = 0;
  dp[1][1][1][1] = 1;
  dp[1][1][2][1] = 0;
  dp[1][2][1][1] = 0;
  dp[1][2][2][1] = 1;
  dp[1][1][1][2] = 0;
  dp[1][1][2][2] = 1;
  dp[1][2][1][2] = 1;
  dp[1][2][2][2] = 0;
  for (int j = 2; j <= n; j++) {
    for (int m = 1; m <= 2 * j && m <= k; m++) {
      dp[j][1][1][m] = dp[j - 1][1][2][m] + dp[j - 1][2][2][m - 1] +
                       dp[j - 1][2][1][m] + dp[j - 1][1][1][m];
      dp[j][1][1][m] %= mod;
      if (m >= 2)
        dp[j][1][2][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m] +
                         dp[j - 1][2][2][m - 1] + dp[j - 1][2][1][m - 2];
      else
        dp[j][1][2][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m] +
                         dp[j - 1][2][2][m - 1];
      dp[j][1][2][m] %= mod;
      if (m >= 2)
        dp[j][2][1][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][2][1][m] +
                         dp[j - 1][1][2][m - 2] + dp[j - 1][2][2][m - 1];
      else
        dp[j][2][1][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][2][1][m] +
                         dp[j - 1][2][2][m - 1];
      dp[j][2][1][m] %= mod;
      dp[j][2][2][m] = dp[j - 1][2][1][m] + dp[j - 1][2][2][m] +
                       dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m];
      dp[j][2][2][m] %= mod;
    }
  }
  long long ans =
      dp[n][1][1][k] + dp[n][1][2][k] + dp[n][2][1][k] + dp[n][2][2][k];
  ans %= mod;
  printf("%I64d", ans);
  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>
const int mod = 998244353;
long long dp[1005][3][3][2005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[1][1][1][0] = 0;
  dp[1][1][2][0] = 0;
  dp[1][2][1][0] = 0;
  dp[1][2][2][0] = 0;
  dp[1][1][1][1] = 1;
  dp[1][1][2][1] = 0;
  dp[1][2][1][1] = 0;
  dp[1][2][2][1] = 1;
  dp[1][1][1][2] = 0;
  dp[1][1][2][2] = 1;
  dp[1][2][1][2] = 1;
  dp[1][2][2][2] = 0;
  for (int j = 2; j <= n; j++) {
    for (int m = 1; m <= 2 * j && m <= k; m++) {
      dp[j][1][1][m] = dp[j - 1][1][2][m] + dp[j - 1][2][2][m - 1] +
                       dp[j - 1][2][1][m] + dp[j - 1][1][1][m];
      dp[j][1][1][m] %= mod;
      if (m >= 2)
        dp[j][1][2][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m] +
                         dp[j - 1][2][2][m - 1] + dp[j - 1][2][1][m - 2];
      else
        dp[j][1][2][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m] +
                         dp[j - 1][2][2][m - 1];
      dp[j][1][2][m] %= mod;
      if (m >= 2)
        dp[j][2][1][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][2][1][m] +
                         dp[j - 1][1][2][m - 2] + dp[j - 1][2][2][m - 1];
      else
        dp[j][2][1][m] = dp[j - 1][1][1][m - 1] + dp[j - 1][2][1][m] +
                         dp[j - 1][2][2][m - 1];
      dp[j][2][1][m] %= mod;
      dp[j][2][2][m] = dp[j - 1][2][1][m] + dp[j - 1][2][2][m] +
                       dp[j - 1][1][1][m - 1] + dp[j - 1][1][2][m];
      dp[j][2][2][m] %= mod;
    }
  }
  long long ans =
      dp[n][1][1][k] + dp[n][1][2][k] + dp[n][2][1][k] + dp[n][2][2][k];
  ans %= mod;
  printf("%I64d", ans);
  return 0;
}

```