#include <bits/stdc++.h>
const int N = 1010;
const int moder = 998244353;
int dp[N][N << 1][4];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[0][1][0] = dp[0][2][1] = dp[0][2][2] = dp[0][1][3] = 1;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < N << 1; ++j) {
      for (int k = 0; k < 4; ++k) {
        if (!dp[i - 1][j][k]) continue;
        for (int u = 0; u < 4; ++u) {
          int add = __builtin_popcount(k ^ u);
          if ((u == 0 || u == 3) && add) {
            --add;
          }
          dp[i][j + add][u] += dp[i - 1][j][k];
          dp[i][j + add][u] -= dp[i][j + add][u] >= moder ? moder : 0;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 4; ++i) {
    ans += dp[n - 1][k][i];
    ans -= ans >= moder ? moder : 0;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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>
const int N = 1010;
const int moder = 998244353;
int dp[N][N << 1][4];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[0][1][0] = dp[0][2][1] = dp[0][2][2] = dp[0][1][3] = 1;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < N << 1; ++j) {
      for (int k = 0; k < 4; ++k) {
        if (!dp[i - 1][j][k]) continue;
        for (int u = 0; u < 4; ++u) {
          int add = __builtin_popcount(k ^ u);
          if ((u == 0 || u == 3) && add) {
            --add;
          }
          dp[i][j + add][u] += dp[i - 1][j][k];
          dp[i][j + add][u] -= dp[i][j + add][u] >= moder ? moder : 0;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 4; ++i) {
    ans += dp[n - 1][k][i];
    ans -= ans >= moder ? moder : 0;
  }
  printf("%d\n", ans);
  return 0;
}

```