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;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int mod = 998244353;
const int maxn = 1010;
long long dp[maxn][maxn * 2][5];
int n, k;
int calc(int a, int b) {
  if (a == 0 || a == 3) {
    if (b == a)
      return 0;
    else
      return 1;
  } else {
    if (b == a)
      return 0;
    else if (b == 3 - a)
      return 2;
    else
      return 0;
  }
}
int main() {
  n = read(), k = read();
  dp[1][1][0] = dp[1][1][3] = 1;
  dp[1][2][1] = dp[1][2][2] = 1;
  for (register int i = (int)(1); i <= (int)(n - 1); i++) {
    for (register int j = (int)(1); j <= (int)(k); j++) {
      for (register int st = (int)(0); st <= (int)(3); st++) {
        if (dp[i][j][st] == 0) continue;
        for (register int nwst = (int)(0); nwst <= (int)(3); nwst++) {
          int nw = j + calc(st, nwst);
          (dp[i + 1][nw][nwst] += dp[i][j][st]) %= mod;
        }
      }
    }
  }
  cout << (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % mod << endl;
  return 0;
}
