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 N = 1e3 + 5;
const int M = 998244353;
const int INF = 2e9 + 5;
const int rx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
const int ry[8] = {0, 0, 1, -1, 1, -1, -1, 1};
long long dp[N][2 * N][4];
int f(int x, int y) {
  if (x == 0 && y == 0) return 0;
  if (x == 0 && y == 1) return 1;
  if (x == 0 && y == 2) return 1;
  if (x == 0 && y == 3) return 1;
  if (x == 1 && y == 0) return 0;
  if (x == 1 && y == 1) return 0;
  if (x == 1 && y == 2) return 2;
  if (x == 1 && y == 3) return 0;
  if (x == 2 && y == 0) return 0;
  if (x == 2 && y == 1) return 2;
  if (x == 2 && y == 2) return 0;
  if (x == 2 && y == 3) return 0;
  if (x == 3 && y == 0) return 1;
  if (x == 3 && y == 1) return 1;
  if (x == 3 && y == 2) return 1;
  if (x == 3 && y == 3) return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  dp[0][1][0] = 1;
  dp[0][1][3] = 1;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  for (int i = 0; i < n - 1; i++) {
    for (int j = 1; j <= k; j++) {
      for (int z = 0; z < 4; z++) {
        for (int x = 0; x < 4; x++) {
          dp[i + 1][j + f(z, x)][x] += dp[i][j][z];
          dp[i + 1][j + f(z, x)][x] %= M;
        }
      }
    }
  }
  long long res = 0;
  for (int i = 0; i < 4; i++) {
    res += dp[n - 1][k][i];
    res %= M;
  }
  cout << res;
  return 0;
};
