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>
int bit(int i, int l) { return (i >> l & 1); }
int dp[1001][2001][4];
signed main() {
  int m = 4, n, k;
  scanf("%d %d", &n, &k);
  std::vector<std::vector<int>> cnt(m, std::vector<int>(m, 0));
  cnt[0][1] = 1;
  cnt[1][0] = 0;
  cnt[0][2] = 1;
  cnt[2][0] = 0;
  cnt[0][3] = 1;
  cnt[3][0] = 1;
  cnt[1][2] = 2;
  cnt[2][1] = 2;
  cnt[1][3] = 0;
  cnt[3][1] = 1;
  cnt[2][3] = 0;
  cnt[3][2] = 1;
  dp[1][1][0] = 1;
  dp[1][1][3] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  const int MOD = 998244353;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      for (int mask1 = 0; mask1 < m; mask1++) {
        if (dp[i][j][mask1] == 0) {
          continue;
        }
        for (int mask2 = 0; mask2 < m; mask2++) {
          if (j + cnt[mask1][mask2] <= k) {
            int &val = dp[i + 1][j + cnt[mask1][mask2]][mask2];
            val += dp[i][j][mask1];
            val %= MOD;
          }
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < m; i++) {
    ans = (ans + dp[n][k][i]) % MOD;
  }
  printf("%d", ans);
}
