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 mod = 998244353;
long long dp[1001][2001][4];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  dp[0][0][0] = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= 2 * i; j++) {
      for (long long last = 0; last < 4; last++) {
        for (long long cur = 0; cur < 4; cur++) {
          long long new_c = 0;
          if (i == 0) {
            new_c = cur == 0 || cur == 3 ? 1 : 2;
          } else {
            if (last == 0 || last == 3) {
              new_c = cur == last ? 0 : 1;
            } else {
              new_c = cur + last == 3 ? 2 : 0;
            }
          }
          if (j + new_c <= k) {
            dp[i + 1][j + new_c][cur] += dp[i][j][last];
            dp[i + 1][j + new_c][cur] %= mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (long long last = 0; last < 4; last++) {
    ans += dp[n][k][last];
    ans %= mod;
  }
  cout << ans << endl;
}
