#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int maxn = 1100;
const int maxk = 2 * maxn;
long long dp[maxn][maxk][4];
long long st[4][4] = {{0, 1, 1, 1}, {1, 0, 1, 1}, {0, 0, 0, 2}, {0, 0, 2, 0}};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[1][1][0] = 1;
  dp[1][1][1] = 1;
  dp[1][2][2] = 1;
  dp[1][2][3] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= k; j++)
      for (int a = 0; a < 4; a++)
        for (int b = 0; b < 4; b++)
          dp[i][j + st[b][a]][a] += dp[i - 1][j][b] % mod;
  printf("%d\n", (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % mod);
  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 long long mod = 998244353;
const int maxn = 1100;
const int maxk = 2 * maxn;
long long dp[maxn][maxk][4];
long long st[4][4] = {{0, 1, 1, 1}, {1, 0, 1, 1}, {0, 0, 0, 2}, {0, 0, 2, 0}};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[1][1][0] = 1;
  dp[1][1][1] = 1;
  dp[1][2][2] = 1;
  dp[1][2][3] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= k; j++)
      for (int a = 0; a < 4; a++)
        for (int b = 0; b < 4; b++)
          dp[i][j + st[b][a]][a] += dp[i - 1][j][b] % mod;
  printf("%d\n", (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % mod);
  return 0;
}

```