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>
#include <bits/stdc++.h>
using namespace std;
template <typename t>
t in(t q) {
  cin >> q;
  return q;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < ((int)size(v)); ++i) {
    os << v[i];
    if (i != ((int)size(v)) - 1) os << ",";
  }
  os << "]";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto it : v) os << "(" << it.first << ":" << it.second << ")";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(" << v.first << "," << v.second << ")";
  return os;
}
const long double PI = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
inline int rand(int l, int r) {
  return uniform_int_distribution<int>(l, r)(rng);
}
inline long long rand(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
long long a[(1005)][2 * (1005)][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10);
  int n, k;
  cin >> n >> k;
  a[0][1][0] = 1;
  a[0][1][3] = 1;
  a[0][2][1] = 1;
  a[0][2][2] = 1;
  for (auto i = 0; i < n; i++) {
    for (auto j = 1; j < k + 1; j++) {
      for (auto l = 0; l < 4; l++) {
        a[i][j][l] %= 998244353ll;
      }
      if (a[i][j][0]) {
        a[i + 1][j][0] += a[i][j][0];
        a[i + 1][j + 1][1] += a[i][j][0];
        a[i + 1][j + 1][2] += a[i][j][0];
        a[i + 1][j + 1][3] += a[i][j][0];
      }
      if (a[i][j][1]) {
        a[i + 1][j][0] += a[i][j][1];
        a[i + 1][j][1] += a[i][j][1];
        a[i + 1][j + 2][2] += a[i][j][1];
        a[i + 1][j][0] += a[i][j][1];
      }
      if (a[i][j][2]) {
        a[i + 1][j][0] += a[i][j][2];
        a[i + 1][j + 2][1] += a[i][j][2];
        a[i + 1][j][2] += a[i][j][2];
        a[i + 1][j][0] += a[i][j][2];
      }
      if (a[i][j][3]) {
        a[i + 1][j + 1][0] += a[i][j][3];
        a[i + 1][j + 1][1] += a[i][j][3];
        a[i + 1][j + 1][2] += a[i][j][3];
        a[i + 1][j][3] += a[i][j][3];
      }
    }
  }
  cout << (a[n - 1][k][0] + a[n - 1][k][1] + a[n - 1][k][2] + a[n - 1][k][3]) %
              998244353ll;
  return 0;
}
