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>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const int maxn = 1000 + 7;
const long long modder = 998244353;
long long dp[maxn][maxn * 2][4];
void adder(long long& x, long long y) {
  x += y;
  x %= modder;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int N, K;
  cin >> N >> K;
  long long transfer[4][4] = {
      {0, 1, 1, 1}, {0, 0, 2, 0}, {0, 2, 0, 0}, {1, 1, 1, 0}};
  dp[1][1][0] = 1LL;
  dp[1][2][1] = 1LL;
  dp[1][2][2] = 1LL;
  dp[1][1][3] = 1LL;
  for (int i = 1; i < N; i++) {
    for (int k = 1; k <= 2 * i; k++) {
      for (int s = 0; s < 4; s++) {
        for (int t = 0; t < 4; t++) {
          adder(dp[i + 1][k + transfer[s][t]][t], dp[i][k][s]);
        }
      }
    }
  }
  long long ans = 0LL;
  for (int s = 0; s < 4; s++) {
    adder(ans, dp[N][K][s]);
  }
  out(ans);
  return 0;
}
