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 = 1000 + 7;
const int MOD = 998244353;
int dp[N][2 * N][4];
int add(int a, int b) { return (a + b) % MOD; }
bool full(int mask) { return (mask == 0 || mask == 3); }
int get(int mask, int nmask) {
  int cnt = __builtin_popcount(mask ^ nmask);
  if (cnt == 0) return 0;
  if (cnt == 2) return (full(mask) ? 1 : 2);
  return (full(mask) ? 0 : 1);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < int(4); i++) dp[1][0][i] = 1;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < int(k + 1); j++) {
      for (int mask = 0; mask < int(4); mask++) {
        for (int nmask = 0; nmask < int(4); nmask++) {
          dp[i + 1][j + get(mask, nmask)][nmask] =
              add(dp[i + 1][j + get(mask, nmask)][nmask], dp[i][j][mask]);
        }
      }
    }
  }
  int ans = 0;
  for (int mask = 0; mask < int(4); mask++) {
    int nw = get(mask, mask ^ 3);
    if (k >= nw) ans = add(ans, dp[n][k - nw][mask]);
  }
  printf("%d\n", ans);
}
