#include <bits/stdc++.h>
using namespace std;
long long get(long long newmask, long long curmask) {
  if (newmask == 0) {
    if (curmask == 0) {
      return (0);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 1) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (2);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 2) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (2);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 3) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (0);
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long i;
  long long dp[n + 1][k + 4][4];
  long long j;
  long long p;
  long long u = 0;
  memset(dp, 0, sizeof(dp));
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (i = 2; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      for (p = 0; p < 4; p++) {
        for (u = 0; u < 4; u++) {
          dp[i][j + get(u, p)][u] += dp[i - 1][j][p];
          dp[i][j + get(u, p)][u] %= 998244353;
        }
      }
    }
  }
  long long sum = 0;
  for (i = 0; i < 4; i++) {
    sum += dp[n][k][i];
    sum %= 998244353;
  }
  cout << sum << "\n";
}

### 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>
using namespace std;
long long get(long long newmask, long long curmask) {
  if (newmask == 0) {
    if (curmask == 0) {
      return (0);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 1) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (2);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 2) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (2);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (1);
    }
  }
  if (newmask == 3) {
    if (curmask == 0) {
      return (1);
    }
    if (curmask == 1) {
      return (0);
    }
    if (curmask == 2) {
      return (0);
    }
    if (curmask == 3) {
      return (0);
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long i;
  long long dp[n + 1][k + 4][4];
  long long j;
  long long p;
  long long u = 0;
  memset(dp, 0, sizeof(dp));
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (i = 2; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      for (p = 0; p < 4; p++) {
        for (u = 0; u < 4; u++) {
          dp[i][j + get(u, p)][u] += dp[i - 1][j][p];
          dp[i][j + get(u, p)][u] %= 998244353;
        }
      }
    }
  }
  long long sum = 0;
  for (i = 0; i < 4; i++) {
    sum += dp[n][k][i];
    sum %= 998244353;
  }
  cout << sum << "\n";
}

```