#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 1, 0, 1};
int dy[] = {0, 1, 1, 0};
int n, k, tmp, dp[1003][2][2][2 * 1003];
int add(int a, int b, int pos) {
  int x = dx[pos];
  int y = dy[pos];
  if (a == 0 && b == 1 && (x == 1 && y == 0)) return 2;
  if (a == 1 && b == 0 && (x == 0 && y == 1)) return 2;
  if (a == 0 && b == 0 && (x == 1 || y == 1)) return 1;
  if (a == 1 && b == 1 && (x == 0 || y == 0)) return 1;
  return 0;
}
int call(int col, int a, int b, int cnt) {
  if (col >= n) {
    if (k == cnt) return 1;
    return 0;
  }
  if (dp[col][a][b][cnt] != -1) return dp[col][a][b][cnt];
  int sum = 0;
  for (int i = 0; i < 4; i++) {
    tmp = add(a, b, i);
    if (tmp + cnt <= k) {
      sum = (sum % 998244353 + call(col + 1, dx[i], dy[i], cnt + tmp)) %
            998244353;
    }
  }
  return dp[col][a][b][cnt] = sum % 998244353;
}
int main() {
  scanf("%d %d", &n, &k);
  memset(dp, -1, sizeof(dp));
  int ans = call(1, 0, 0, 1) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 1, 1, 1)) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 1, 0, 2)) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 0, 1, 2)) % 998244353;
  printf("%d\n", ans);
}

### 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;
int dx[] = {0, 1, 0, 1};
int dy[] = {0, 1, 1, 0};
int n, k, tmp, dp[1003][2][2][2 * 1003];
int add(int a, int b, int pos) {
  int x = dx[pos];
  int y = dy[pos];
  if (a == 0 && b == 1 && (x == 1 && y == 0)) return 2;
  if (a == 1 && b == 0 && (x == 0 && y == 1)) return 2;
  if (a == 0 && b == 0 && (x == 1 || y == 1)) return 1;
  if (a == 1 && b == 1 && (x == 0 || y == 0)) return 1;
  return 0;
}
int call(int col, int a, int b, int cnt) {
  if (col >= n) {
    if (k == cnt) return 1;
    return 0;
  }
  if (dp[col][a][b][cnt] != -1) return dp[col][a][b][cnt];
  int sum = 0;
  for (int i = 0; i < 4; i++) {
    tmp = add(a, b, i);
    if (tmp + cnt <= k) {
      sum = (sum % 998244353 + call(col + 1, dx[i], dy[i], cnt + tmp)) %
            998244353;
    }
  }
  return dp[col][a][b][cnt] = sum % 998244353;
}
int main() {
  scanf("%d %d", &n, &k);
  memset(dp, -1, sizeof(dp));
  int ans = call(1, 0, 0, 1) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 1, 1, 1)) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 1, 0, 2)) % 998244353;
  memset(dp, -1, sizeof(dp));
  ans = (ans + call(1, 0, 1, 2)) % 998244353;
  printf("%d\n", ans);
}

```