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;
int read(int x = 0) {
  int f = 1, c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  return f * x;
}
long long N, K, f[2010][2010][5], cnt[5][5];
int main() {
  long long i, j, k, l, ans = 0;
  N = read(), K = read();
  f[1][1][0] = f[1][2][1] = f[1][2][2] = f[1][1][3] = 1;
  cnt[0][0] = 0;
  cnt[0][1] = 1;
  cnt[0][2] = 1;
  cnt[0][3] = 1;
  cnt[1][0] = 0;
  cnt[1][1] = 0;
  cnt[1][2] = 2;
  cnt[1][3] = 0;
  cnt[2][0] = 0;
  cnt[2][1] = 2;
  cnt[2][3] = 0;
  cnt[3][0] = 1;
  cnt[3][1] = 1;
  cnt[3][2] = 1;
  cnt[3][3] = 0;
  for (i = 1; i < N; i++)
    for (j = 1; j <= (i << 1); j++)
      for (k = 0; k < 4; k++)
        for (l = 0; l < 4; l++) {
          f[i + 1][j + cnt[k][l]][l] =
              (f[i + 1][j + cnt[k][l]][l] + f[i][j][k]) % 998244353ll;
        }
  for (k = 0; k < 4; k++) ans = (ans + f[N][K][k]) % 998244353ll;
  printf("%I64d\n", ans);
  return 0;
}
