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;
const int N = 100005;
const int P = 1000000007;
const long long mod = 998244353;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long double eps = 1e-8;
long long qmod(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    assert(ans > 0);
    if (n & 1ll) ans = ans * a % mod;
    n >>= 1;
    a = a * a % mod;
  }
  return ans % mod;
}
long long dp[1005][2005][5];
int n, k;
int main() {
  int n, k;
  cin >> n >> k;
  memset(dp, 0, sizeof(dp));
  dp[1][1][0] = 1;
  dp[1][1][3] = 1;
  dp[1][2][2] = 1;
  dp[1][2][1] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= 2 * n; ++j) {
      dp[i][j][0] += dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j - 1][3];
      dp[i][j][3] += dp[i - 1][j][3] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j - 1][0];
      dp[i][j][1] += dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                     dp[i - 1][j][1] + (j >= 2 ? dp[i - 1][j - 2][2] : 0);
      dp[i][j][2] += dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                     dp[i - 1][j][2] + (j >= 2 ? dp[i - 1][j - 2][1] : 0);
      for (int k = 0; k < 4; ++k) dp[i][j][k] %= mod;
    }
  }
  long long ans = 0;
  for (int i = 0; i < 4; ++i) ans = ans + dp[n][k][i], ans %= mod;
  cout << ans << endl;
  return 0;
}
