#include <bits/stdc++.h>
using namespace std;
int n, m;
long long dp[1010][2010][5];
void precal() {
  char sa[4][2] = {'0', '0', '0', '1', '1', '0', '1', '1'};
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 1; i < 1000; i++) {
    for (int j = 1; j <= 2000; j++) {
      for (int k1 = 0; k1 < 4; k1++) {
        for (int k2 = 0; k2 < 4; k2++) {
          int themj = 0;
          themj += sa[k1][0] == sa[k2][0] ? 0 : 1;
          themj += sa[k1][1] == sa[k2][1] ? 0 : 1;
          themj -= sa[k2][0] == sa[k2][1] ? 1 : 0;
          themj = max(0, themj);
          dp[i + 1][j + themj][k2] += dp[i][j][k1];
          dp[i + 1][j + themj][k2] %= 998244353;
        }
      }
    }
  }
}
void solve() {}
void output() {
  printf("%I64d\n",
         (dp[n][m][0] + dp[n][m][1] + dp[n][m][2] + dp[n][m][3]) % 998244353);
}
int main() {
  precal();
  while (scanf("%d %d", &n, &m) > 0) {
    output();
  }
  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;
int n, m;
long long dp[1010][2010][5];
void precal() {
  char sa[4][2] = {'0', '0', '0', '1', '1', '0', '1', '1'};
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 1; i < 1000; i++) {
    for (int j = 1; j <= 2000; j++) {
      for (int k1 = 0; k1 < 4; k1++) {
        for (int k2 = 0; k2 < 4; k2++) {
          int themj = 0;
          themj += sa[k1][0] == sa[k2][0] ? 0 : 1;
          themj += sa[k1][1] == sa[k2][1] ? 0 : 1;
          themj -= sa[k2][0] == sa[k2][1] ? 1 : 0;
          themj = max(0, themj);
          dp[i + 1][j + themj][k2] += dp[i][j][k1];
          dp[i + 1][j + themj][k2] %= 998244353;
        }
      }
    }
  }
}
void solve() {}
void output() {
  printf("%I64d\n",
         (dp[n][m][0] + dp[n][m][1] + dp[n][m][2] + dp[n][m][3]) % 998244353);
}
int main() {
  precal();
  while (scanf("%d %d", &n, &m) > 0) {
    output();
  }
  return 0;
}

```