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 = 1001, M = 3;
const int MOD = 998244353;
int dp[N][N * 2][M][M];
int get(int u1, int d1, int u2, int d2) {
  int res = 0;
  if ((u2 == d2 && (u1 == u2 || d1 == d2))) {
    return 0;
  }
  if (u1 == u2 && d1 == d2) {
    return 0;
  }
  if (u1 != u2 && d1 != d2 && d2 != u2) {
    return 2;
  }
  return 1;
}
void add(int &a, int &b) { a = (a % MOD + b % MOD) % MOD; }
int main() {
  int n, k;
  cin >> n >> k;
  ;
  dp[1][2][1][0] = 1;
  dp[1][1][1][1] = 1;
  dp[1][2][0][1] = 1;
  dp[1][1][0][0] = 1;
  for (int i = 1; i < n; i++) {
    for (int cost = 0; cost <= k; cost++) {
      for (int u1 = 0; u1 <= 1; u1++) {
        for (int d1 = 0; d1 <= 1; d1++) {
          for (int u2 = 0; u2 <= 1; u2++) {
            for (int d2 = 0; d2 <= 1; d2++) {
              add(dp[i + 1][cost + get(u1, d1, u2, d2)][u2][d2],
                  dp[i][cost][u1][d1]);
            }
          }
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= 1; i++) {
    for (int j = 0; j <= 1; j++) {
      add(ans, dp[n][k][i][j]);
    }
  }
  cout << ans << endl;
}
