#include <bits/stdc++.h>
using namespace std;
const int inf = 998244353;
struct str {
  long long one_u, one_d, two, zero;
  str() {
    one_u = 0;
    one_d = 0;
    two = 0;
    zero = 0;
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long kol_vo, k;
  cin >> kol_vo >> k;
  vector<vector<str> > dp(kol_vo, vector<str>(k + 2));
  dp[0][2].one_u = 1;
  dp[0][2].one_d = 1;
  dp[0][1].zero = 1;
  dp[0][1].two = 1;
  for (long long c = 1; c < kol_vo; c++) {
    for (long long nowk = 1; nowk <= k; nowk++) {
      dp[c][nowk].zero = dp[c - 1][nowk].one_u + dp[c - 1][nowk].one_d +
                         dp[c - 1][nowk].zero + dp[c - 1][nowk - 1].two;
      dp[c][nowk].two = dp[c - 1][nowk].one_u + dp[c - 1][nowk].one_d +
                        dp[c - 1][nowk].two + dp[c - 1][nowk - 1].zero;
      dp[c][nowk].one_u = dp[c - 1][nowk].one_u + dp[c - 1][nowk - 1].two +
                          dp[c - 1][nowk - 1].zero;
      dp[c][nowk].one_d = dp[c - 1][nowk].one_d + dp[c - 1][nowk - 1].two +
                          dp[c - 1][nowk - 1].zero;
      if (nowk >= 2) dp[c][nowk].one_u += dp[c - 1][nowk - 2].one_d;
      if (nowk >= 2) dp[c][nowk].one_d += dp[c - 1][nowk - 2].one_u;
      dp[c][nowk].one_u %= inf;
      dp[c][nowk].one_d %= inf;
      dp[c][nowk].two %= inf;
      dp[c][nowk].zero %= inf;
    }
  }
  long long answ = 0;
  answ += dp[kol_vo - 1][k].one_u + dp[kol_vo - 1][k].zero +
          dp[kol_vo - 1][k].two + dp[kol_vo - 1][k].one_d;
  cout << answ % inf;
  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>
using namespace std;
const int inf = 998244353;
struct str {
  long long one_u, one_d, two, zero;
  str() {
    one_u = 0;
    one_d = 0;
    two = 0;
    zero = 0;
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long kol_vo, k;
  cin >> kol_vo >> k;
  vector<vector<str> > dp(kol_vo, vector<str>(k + 2));
  dp[0][2].one_u = 1;
  dp[0][2].one_d = 1;
  dp[0][1].zero = 1;
  dp[0][1].two = 1;
  for (long long c = 1; c < kol_vo; c++) {
    for (long long nowk = 1; nowk <= k; nowk++) {
      dp[c][nowk].zero = dp[c - 1][nowk].one_u + dp[c - 1][nowk].one_d +
                         dp[c - 1][nowk].zero + dp[c - 1][nowk - 1].two;
      dp[c][nowk].two = dp[c - 1][nowk].one_u + dp[c - 1][nowk].one_d +
                        dp[c - 1][nowk].two + dp[c - 1][nowk - 1].zero;
      dp[c][nowk].one_u = dp[c - 1][nowk].one_u + dp[c - 1][nowk - 1].two +
                          dp[c - 1][nowk - 1].zero;
      dp[c][nowk].one_d = dp[c - 1][nowk].one_d + dp[c - 1][nowk - 1].two +
                          dp[c - 1][nowk - 1].zero;
      if (nowk >= 2) dp[c][nowk].one_u += dp[c - 1][nowk - 2].one_d;
      if (nowk >= 2) dp[c][nowk].one_d += dp[c - 1][nowk - 2].one_u;
      dp[c][nowk].one_u %= inf;
      dp[c][nowk].one_d %= inf;
      dp[c][nowk].two %= inf;
      dp[c][nowk].zero %= inf;
    }
  }
  long long answ = 0;
  answ += dp[kol_vo - 1][k].one_u + dp[kol_vo - 1][k].zero +
          dp[kol_vo - 1][k].two + dp[kol_vo - 1][k].one_d;
  cout << answ % inf;
  return 0;
}

```