#include <bits/stdc++.h>
using namespace std;
long long n, k;
vector<long long> v;
long long dp[1001][2001][4];
long long rec(long long i, long long mask, long long comp) {
  if (i == n && comp == k)
    return 1;
  else if (i == n)
    return 0;
  else if (comp > k)
    return 0;
  if (dp[i][comp][mask] != -1) return dp[i][comp][mask];
  if (i == 0) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp + 1) + rec(i + 1, 1, comp + 2) +
                rec(i + 1, 2, comp + 2) + rec(i + 1, 3, comp + 1)) %
               998244353;
  }
  if (mask == 0) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp) + rec(i + 1, 3, comp + 1) +
                rec(i + 1, 1, comp + 1) + rec(i + 1, 2, comp + 1)) %
               998244353;
  } else if (mask == 1) {
    return dp[i][comp][mask] = (rec(i + 1, 0, comp) + rec(i + 1, 1, comp) +
                                rec(i + 1, 2, comp + 2) + rec(i + 1, 3, comp)) %
                               998244353;
  } else if (mask == 2) {
    return dp[i][comp][mask] = (rec(i + 1, 0, comp) + rec(i + 1, 1, comp + 2) +
                                rec(i + 1, 2, comp) + rec(i + 1, 3, comp)) %
                               998244353;
  } else if (mask == 3) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp + 1) + rec(i + 1, 1, comp + 1) +
                rec(i + 1, 2, comp + 1) + rec(i + 1, 3, comp)) %
               998244353;
  }
}
void solve() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  cout << rec(0, 0, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 1;
  while (q--) {
    v.clear();
    solve();
  }
}

### Prompt

Develop a solution in cpp 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 n, k;
vector<long long> v;
long long dp[1001][2001][4];
long long rec(long long i, long long mask, long long comp) {
  if (i == n && comp == k)
    return 1;
  else if (i == n)
    return 0;
  else if (comp > k)
    return 0;
  if (dp[i][comp][mask] != -1) return dp[i][comp][mask];
  if (i == 0) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp + 1) + rec(i + 1, 1, comp + 2) +
                rec(i + 1, 2, comp + 2) + rec(i + 1, 3, comp + 1)) %
               998244353;
  }
  if (mask == 0) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp) + rec(i + 1, 3, comp + 1) +
                rec(i + 1, 1, comp + 1) + rec(i + 1, 2, comp + 1)) %
               998244353;
  } else if (mask == 1) {
    return dp[i][comp][mask] = (rec(i + 1, 0, comp) + rec(i + 1, 1, comp) +
                                rec(i + 1, 2, comp + 2) + rec(i + 1, 3, comp)) %
                               998244353;
  } else if (mask == 2) {
    return dp[i][comp][mask] = (rec(i + 1, 0, comp) + rec(i + 1, 1, comp + 2) +
                                rec(i + 1, 2, comp) + rec(i + 1, 3, comp)) %
                               998244353;
  } else if (mask == 3) {
    return dp[i][comp][mask] =
               (rec(i + 1, 0, comp + 1) + rec(i + 1, 1, comp + 1) +
                rec(i + 1, 2, comp + 1) + rec(i + 1, 3, comp)) %
               998244353;
  }
}
void solve() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  cout << rec(0, 0, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 1;
  while (q--) {
    v.clear();
    solve();
  }
}

```