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 maxn = 1e3 + 5;
const int MOD = 998244353;
long long dp[maxn][maxn << 1][4];
int main() {
  dp[1][1][0] = 1;
  dp[1][1][3] = 1;
  dp[1][2][2] = 1;
  dp[1][2][1] = 1;
  for (int i = 2; i <= 1000; i++) {
    for (int j = 1; j <= i * 2; j++) {
      for (int k = 0; k <= 3; k++) {
        dp[i][j][k] += dp[i - 1][j][k];
        if (k == 0 || k == 3) {
          dp[i][j][k] += dp[i - 1][j][k ^ 1];
          dp[i][j][k] += dp[i - 1][j][k ^ 2];
        }
        if (k == 0 || k == 3) {
          dp[i][j][k] += dp[i - 1][j - 1][k ^ k];
        } else {
          dp[i][j][k] += dp[i - 1][j - 1][0];
          dp[i][j][k] += dp[i - 1][j - 1][3];
        }
        if (k == 1 || k == 2) {
          if (j >= 2) dp[i][j][k] += dp[i - 1][j - 2][k ^ 3];
        }
        dp[i][j][k] %= MOD;
      }
    }
  }
  int N, K;
  while (cin >> N >> K) {
    long long ans = 0;
    for (int sta = 0; sta <= 3; sta++) ans += dp[N][K][sta];
    ans %= MOD;
    cout << ans << endl;
  }
}
