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;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / GCD(a, b); }
const long long N = 1e3 + 7;
const long long inf = 1e18 + 1;
const long long mod = 998244353;
const double eps = 1e-8;
int n, k;
int dp[1001][4][2011];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  memset(dp, 0, sizeof dp);
  dp[1][0][1] = 1;
  dp[1][1][1] = 1;
  dp[1][2][2] = 1;
  dp[1][3][2] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= min(k, 2 * i); j++) {
      dp[i][0][j] = (dp[i - 1][0][j] + dp[i - 1][1][j - 1] % mod +
                     dp[i - 1][2][j] % mod + dp[i - 1][3][j] % mod) %
                    mod;
      dp[i][1][j] = (dp[i - 1][0][j] + dp[i - 1][1][j - 1] % mod +
                     dp[i - 1][2][j] % mod + dp[i - 1][3][j] % mod) %
                    mod;
      dp[i][2][j] = (dp[i - 1][0][j - 1] + dp[i - 1][1][j - 1] % mod +
                     dp[i - 1][2][j] % mod) %
                    mod;
      dp[i][3][j] = (dp[i - 1][0][j - 1] + dp[i - 1][1][j - 1] % mod +
                     dp[i - 1][3][j] % mod) %
                    mod;
      if (j - 2 >= 0)
        dp[i][2][j] = (dp[i][2][j] + dp[i - 1][3][j - 2]) % mod,
        dp[i][3][j] = (dp[i][3][j] + dp[i - 1][2][j - 2]) % mod;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= 3; i++) ans = (ans + dp[n][i][k]) % mod;
  cout << ans << "\n";
  return 0;
}
