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 mod = 998244353;
int n, k;
int f[1010][2010][4];
int ans;
void add(int &a, int b) { a = (a + b) % mod; }
int get(int mask, int nmask) {
  int cnt = __builtin_popcount(mask ^ nmask);
  if (cnt == 0) return 0;
  if (cnt == 2) return (((mask) == 0 || (mask) == 3) ? 1 : 2);
  return (((mask) == 0 || (mask) == 3) ? 0 : 1);
}
int main() {
  cin >> n >> k;
  int i, j, l;
  for (i = 0; i < 4; i++) f[1][0][i] = 1;
  for (i = 1; i < n; i++) {
    for (j = 0; j <= k; j++) {
      for (int mask = 0; mask < 4; mask++) {
        for (int nmask = 0; nmask < 4; nmask++)
          add(f[i + 1][j + get(mask, nmask)][nmask], f[i][j][mask]);
      }
    }
  }
  for (int mask = 0; mask < 4; mask++) {
    int nw = get(mask, mask ^ 3);
    if (k >= nw) add(ans, f[n][k - nw][mask]);
  }
  cout << ans << endl;
  return 0;
}
