#include <bits/stdc++.h>
using namespace std;
const long long N = 1000 + 10;
const long long M = 3e4 + 10;
const long long inf = 1e8 + 7;
const long long Mod = 998244353;
long long dp[N][4][2 * N];
long long n, k;
int main() {
  scanf("%lld%lld", &n, &k);
  dp[1][0][1] = 1;
  dp[1][1][2] = 1;
  dp[1][2][2] = 1;
  dp[1][3][1] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= k; ++j) {
      if (dp[i - 1][0][j] > 0) {
        (dp[i][0][j] += dp[i - 1][0][j]) %= Mod;
        (dp[i][1][j + 1] += dp[i - 1][0][j]) %= Mod;
        (dp[i][2][j + 1] += dp[i - 1][0][j]) %= Mod;
        (dp[i][3][j + 1] += dp[i - 1][0][j]) %= Mod;
      }
      if (dp[i - 1][1][j] > 0) {
        (dp[i][0][j] += dp[i - 1][1][j]) %= Mod;
        (dp[i][1][j] += dp[i - 1][1][j]) %= Mod;
        (dp[i][2][j + 2] += dp[i - 1][1][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][1][j]) %= Mod;
      }
      if (dp[i - 1][2][j] > 0) {
        (dp[i][0][j] += dp[i - 1][2][j]) %= Mod;
        (dp[i][1][j + 2] += dp[i - 1][2][j]) %= Mod;
        (dp[i][2][j] += dp[i - 1][2][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][2][j]) %= Mod;
      }
      if (dp[i - 1][3][j] > 0) {
        (dp[i][0][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][1][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][2][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][3][j]) %= Mod;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < 4; ++i) {
    (ans += dp[n][i][k]) %= Mod;
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Generate 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 long long N = 1000 + 10;
const long long M = 3e4 + 10;
const long long inf = 1e8 + 7;
const long long Mod = 998244353;
long long dp[N][4][2 * N];
long long n, k;
int main() {
  scanf("%lld%lld", &n, &k);
  dp[1][0][1] = 1;
  dp[1][1][2] = 1;
  dp[1][2][2] = 1;
  dp[1][3][1] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= k; ++j) {
      if (dp[i - 1][0][j] > 0) {
        (dp[i][0][j] += dp[i - 1][0][j]) %= Mod;
        (dp[i][1][j + 1] += dp[i - 1][0][j]) %= Mod;
        (dp[i][2][j + 1] += dp[i - 1][0][j]) %= Mod;
        (dp[i][3][j + 1] += dp[i - 1][0][j]) %= Mod;
      }
      if (dp[i - 1][1][j] > 0) {
        (dp[i][0][j] += dp[i - 1][1][j]) %= Mod;
        (dp[i][1][j] += dp[i - 1][1][j]) %= Mod;
        (dp[i][2][j + 2] += dp[i - 1][1][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][1][j]) %= Mod;
      }
      if (dp[i - 1][2][j] > 0) {
        (dp[i][0][j] += dp[i - 1][2][j]) %= Mod;
        (dp[i][1][j + 2] += dp[i - 1][2][j]) %= Mod;
        (dp[i][2][j] += dp[i - 1][2][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][2][j]) %= Mod;
      }
      if (dp[i - 1][3][j] > 0) {
        (dp[i][0][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][1][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][2][j + 1] += dp[i - 1][3][j]) %= Mod;
        (dp[i][3][j] += dp[i - 1][3][j]) %= Mod;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < 4; ++i) {
    (ans += dp[n][i][k]) %= Mod;
  }
  printf("%lld", ans);
  return 0;
}

```