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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
long long gcd(long long x, long long y) {
  if (x > y) swap(x, y);
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long F[1005][1005 * 2][4];
long long n, k;
int main() {
  n = read(), k = read();
  F[1][1][0] = F[1][1][3] = 1;
  F[1][2][1] = F[1][2][2] = 1;
  for (long long i = 2; i <= n; i++) {
    for (long long j = 1; j <= k; j++) {
      if (j >= 0)
        F[i][j][0] =
            (((F[i - 1][j][0] + F[i - 1][j][1]) % 998244353 + F[i - 1][j][2]) %
                 998244353 +
             F[i - 1][j - 1][3]) %
            998244353;
      if (j >= 1)
        F[i][j][1] = (((F[i - 1][j - 1][0] + F[i - 1][j][1]) % 998244353 +
                       (j < 2 ? 0 : F[i - 1][j - 2][2])) %
                          998244353 +
                      F[i - 1][j - 1][3]) %
                     998244353;
      if (j >= 1)
        F[i][j][2] = (((F[i - 1][j - 1][0] + (j < 2 ? 0 : F[i - 1][j - 2][1])) %
                           998244353 +
                       F[i - 1][j][2]) %
                          998244353 +
                      F[i - 1][j - 1][3]) %
                     998244353;
      if (j >= 0)
        F[i][j][3] = (((F[i - 1][j - 1][0] + F[i - 1][j][1]) % 998244353 +
                       F[i - 1][j][2]) %
                          998244353 +
                      F[i - 1][j][3]) %
                     998244353;
    }
  }
  long long ans =
      ((F[n][k][0] + F[n][k][1]) % 998244353 + F[n][k][2]) % 998244353 +
      F[n][k][3];
  ans %= 998244353;
  cout << ans;
  return 0;
}
