#include <bits/stdc++.h>
using namespace std;
const int d[4][4] = {{0, 1, 1, 1}, {0, 0, 2, 0}, {0, 2, 0, 0}, {1, 1, 1, 0}};
const long long mod = 998244353;
int n, K;
long long f[1010][1010 << 1][4], ans;
int main() {
  scanf("%d%d", &n, &K);
  f[1][1][0] = f[1][1][3] = 1;
  f[1][2][1] = f[1][2][2] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 1; j <= 2 * i; j++)
      for (int k = 0; k < 4; k++)
        if (f[i][j][k])
          for (int l = 0; l < 4; l++)
            (f[i + 1][j + d[k][l]][l] += f[i][j][k]) %= mod;
  for (int i = 0; i < 4; i++) (ans += f[n][K][i]) %= mod;
  printf("%lld\n", 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 int d[4][4] = {{0, 1, 1, 1}, {0, 0, 2, 0}, {0, 2, 0, 0}, {1, 1, 1, 0}};
const long long mod = 998244353;
int n, K;
long long f[1010][1010 << 1][4], ans;
int main() {
  scanf("%d%d", &n, &K);
  f[1][1][0] = f[1][1][3] = 1;
  f[1][2][1] = f[1][2][2] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 1; j <= 2 * i; j++)
      for (int k = 0; k < 4; k++)
        if (f[i][j][k])
          for (int l = 0; l < 4; l++)
            (f[i + 1][j + d[k][l]][l] += f[i][j][k]) %= mod;
  for (int i = 0; i < 4; i++) (ans += f[n][K][i]) %= mod;
  printf("%lld\n", ans);
  return 0;
}

```