#include <bits/stdc++.h>
constexpr const int64_t mod = 998244353;
int64_t dp[1005][2005][4];
int n, k;
int main(int argc, char* argv[]) {
  scanf("%d%d", &n, &k);
  std::fill(dp[1][0], dp[1][0] + 4, 1LL);
  for (int i = 2; i <= n; ++i) {
    for (int j = 0; j < k; ++j) {
      dp[i][j][0] = dp[i - 1][j][0];
      if (j >= 1) {
        dp[i][j][0] +=
            dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2] + dp[i - 1][j - 1][3];
        dp[i][j][0] %= mod;
      }
      dp[i][j][1] = dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][3];
      if (j >= 2) dp[i][j][1] += dp[i - 1][j - 2][2];
      dp[i][j][1] %= mod;
      dp[i][j][2] = dp[i - 1][j][0] + dp[i - 1][j][2] + dp[i - 1][j][3];
      if (j >= 2) dp[i][j][2] += dp[i - 1][j - 2][1];
      dp[i][j][2] %= mod;
      dp[i][j][3] = dp[i - 1][j][3];
      if (j >= 1) {
        dp[i][j][3] +=
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
        dp[i][j][3] %= mod;
      }
    }
  }
  int64_t ans = dp[n][k - 1][0] + dp[n][k - 1][3];
  if (k >= 2) ans += dp[n][k - 2][1] + dp[n][k - 2][2];
  ans %= mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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>
constexpr const int64_t mod = 998244353;
int64_t dp[1005][2005][4];
int n, k;
int main(int argc, char* argv[]) {
  scanf("%d%d", &n, &k);
  std::fill(dp[1][0], dp[1][0] + 4, 1LL);
  for (int i = 2; i <= n; ++i) {
    for (int j = 0; j < k; ++j) {
      dp[i][j][0] = dp[i - 1][j][0];
      if (j >= 1) {
        dp[i][j][0] +=
            dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2] + dp[i - 1][j - 1][3];
        dp[i][j][0] %= mod;
      }
      dp[i][j][1] = dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][3];
      if (j >= 2) dp[i][j][1] += dp[i - 1][j - 2][2];
      dp[i][j][1] %= mod;
      dp[i][j][2] = dp[i - 1][j][0] + dp[i - 1][j][2] + dp[i - 1][j][3];
      if (j >= 2) dp[i][j][2] += dp[i - 1][j - 2][1];
      dp[i][j][2] %= mod;
      dp[i][j][3] = dp[i - 1][j][3];
      if (j >= 1) {
        dp[i][j][3] +=
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][2];
        dp[i][j][3] %= mod;
      }
    }
  }
  int64_t ans = dp[n][k - 1][0] + dp[n][k - 1][3];
  if (k >= 2) ans += dp[n][k - 2][1] + dp[n][k - 2][2];
  ans %= mod;
  printf("%lld\n", ans);
  return 0;
}

```