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 powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long _a, long long _b) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long gcd(long long a, long long b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
long long dp[1005][2010][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  long long M = 998244353;
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][1][2] = 1;
  dp[1][2][3] = 1;
  for (long long i = 2; i <= n; i++) {
    for (long long j = 1; j <= k; j++) {
      dp[i][j][0] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j - 1][2] +
                     dp[i - 1][j][3]) %
                    M;
      dp[i][j][1] =
          (dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j - 1][2]) % M;
      if (j >= 2) dp[i][j][1] = (dp[i - 1][j - 2][3] + dp[i][j][1]) % M;
      dp[i][j][2] = (dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j][3]) %
                    M;
      dp[i][j][3] =
          (dp[i - 1][j - 1][0] + dp[i - 1][j - 1][2] + dp[i - 1][j][3]) % M;
      if (j >= 2) dp[i][j][3] = (dp[i - 1][j - 2][1] + dp[i][j][3]) % M;
    }
  }
  long long ans = (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % M;
  cout << ans;
  return 0;
}
