#include <bits/stdc++.h>
using namespace std;
long long dp[1005][2005][5];
long long n, k;
void rset() {
  for (long long i = 0; i < 1005; i++) {
    for (long long j = 0; j < 2005; j++) {
      for (long long k = 0; k < 5; k++) dp[i][j][k] = -1;
    }
  }
}
long long func(long long pos, long long taken, long long pat) {
  if (pos > n) {
    if (taken == k) return 1;
    return 0;
  }
  if (dp[pos][taken][pat] != -1) return dp[pos][taken][pat];
  long long ret = 0;
  if (pat == 0) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 3);
    ret %= 998244353;
  }
  if (pat == 1) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 2, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  if (pat == 2) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 2, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  if (pat == 3) {
    ret += func(pos + 1, taken + 1, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  return dp[pos][taken][pat] = ret;
}
int main() {
  rset();
  cin >> n >> k;
  long long ans = 0;
  ans += func(2, 1, 0);
  ans += func(2, 2, 1);
  ans += func(2, 2, 2);
  ans += func(2, 1, 3);
  ans %= 998244353;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
long long dp[1005][2005][5];
long long n, k;
void rset() {
  for (long long i = 0; i < 1005; i++) {
    for (long long j = 0; j < 2005; j++) {
      for (long long k = 0; k < 5; k++) dp[i][j][k] = -1;
    }
  }
}
long long func(long long pos, long long taken, long long pat) {
  if (pos > n) {
    if (taken == k) return 1;
    return 0;
  }
  if (dp[pos][taken][pat] != -1) return dp[pos][taken][pat];
  long long ret = 0;
  if (pat == 0) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 3);
    ret %= 998244353;
  }
  if (pat == 1) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 2, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  if (pat == 2) {
    ret += func(pos + 1, taken, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 2, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  if (pat == 3) {
    ret += func(pos + 1, taken + 1, 0);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 1);
    ret %= 998244353;
    ret += func(pos + 1, taken + 1, 2);
    ret %= 998244353;
    ret += func(pos + 1, taken, 3);
    ret %= 998244353;
  }
  return dp[pos][taken][pat] = ret;
}
int main() {
  rset();
  cin >> n >> k;
  long long ans = 0;
  ans += func(2, 1, 0);
  ans += func(2, 2, 1);
  ans += func(2, 2, 2);
  ans += func(2, 1, 3);
  ans %= 998244353;
  cout << ans << endl;
  return 0;
}

```