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 INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
const int MOD = 998244353;
int m[4][4];
long long int n, k;
long long int memo[1002][2004][4];
long long int pd(int ind, int comp, int mask) {
  if (comp > k) return 0;
  if (ind == n) {
    if (comp == k) return 1;
    return 0;
  }
  long long int &pdm = memo[ind][comp][mask];
  if (pdm != -1) return pdm;
  pdm = 0;
  for (int i = 0; i < 4; i++) {
    pdm += pd(ind + 1, comp + m[mask][i], i);
    pdm = pdm % MOD;
  }
  return pdm;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  memset(memo, -1, sizeof(memo));
  for (int i = 0; i < 4; i++) m[i][i] = 0;
  m[0][1] = 1;
  m[0][2] = 1;
  m[0][3] = 1;
  m[1][0] = 0;
  m[1][2] = 2;
  m[1][3] = 0;
  m[2][0] = 0;
  m[2][1] = 2;
  m[2][3] = 0;
  m[3][0] = 1;
  m[3][1] = 1;
  m[3][2] = 1;
  long long int ans;
  ans = pd(1, 1, 0);
  ans = ans % MOD;
  ans += pd(1, 2, 1);
  ans = ans % MOD;
  ans += pd(1, 2, 2);
  ans = ans % MOD;
  ans += pd(1, 1, 3);
  ans = ans % MOD;
  cout << ans << endl;
}
