#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
const int K = (1 << 17) + 10;
const long long int INF = 2e18 + 500;
const int inf = 1e9;
const int SQRT = 500;
const int lgN = 20;
const int MOD = 998244353;
long long int dp[1005][2005][4];
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  dp[1][1][0] = 1;
  dp[1][2][1] = dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= min(2 * i, k); j++) {
      for (int state = 0; state < 4; state++) {
        long long v = 0;
        if (state == 0) {
          v += dp[i - 1][j][0];
          v += dp[i - 1][j][1];
          v += dp[i - 1][j][2];
          v += dp[i - 1][j - 1][3];
        } else if (state == 1) {
          v += dp[i - 1][j - 1][0];
          v += dp[i - 1][j][1];
          if (j > 1) {
            v += dp[i - 1][j - 2][2];
          }
          v += dp[i - 1][j - 1][3];
        } else if (state == 2) {
          v += dp[i - 1][j - 1][0];
          if (j > 1) {
            v += dp[i - 1][j - 2][1];
          }
          v += dp[i - 1][j][2];
          v += dp[i - 1][j - 1][3];
        } else {
          v += dp[i - 1][j - 1][0];
          v += dp[i - 1][j][1];
          v += dp[i - 1][j][2];
          v += dp[i - 1][j][3];
        }
        dp[i][j][state] = v % MOD;
        assert(dp[i][j][state] >= 0 && dp[i][j][state] < MOD);
      }
    }
  }
  long long ans = dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3];
  ans %= MOD;
  assert(ans >= 0 && ans <= MOD);
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 N = 2e6 + 5;
const int K = (1 << 17) + 10;
const long long int INF = 2e18 + 500;
const int inf = 1e9;
const int SQRT = 500;
const int lgN = 20;
const int MOD = 998244353;
long long int dp[1005][2005][4];
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  dp[1][1][0] = 1;
  dp[1][2][1] = dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= min(2 * i, k); j++) {
      for (int state = 0; state < 4; state++) {
        long long v = 0;
        if (state == 0) {
          v += dp[i - 1][j][0];
          v += dp[i - 1][j][1];
          v += dp[i - 1][j][2];
          v += dp[i - 1][j - 1][3];
        } else if (state == 1) {
          v += dp[i - 1][j - 1][0];
          v += dp[i - 1][j][1];
          if (j > 1) {
            v += dp[i - 1][j - 2][2];
          }
          v += dp[i - 1][j - 1][3];
        } else if (state == 2) {
          v += dp[i - 1][j - 1][0];
          if (j > 1) {
            v += dp[i - 1][j - 2][1];
          }
          v += dp[i - 1][j][2];
          v += dp[i - 1][j - 1][3];
        } else {
          v += dp[i - 1][j - 1][0];
          v += dp[i - 1][j][1];
          v += dp[i - 1][j][2];
          v += dp[i - 1][j][3];
        }
        dp[i][j][state] = v % MOD;
        assert(dp[i][j][state] >= 0 && dp[i][j][state] < MOD);
      }
    }
  }
  long long ans = dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3];
  ans %= MOD;
  assert(ans >= 0 && ans <= MOD);
  cout << ans << endl;
  return 0;
}

```