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 long long M = 2e3 + 5, mod = 998244353;
long long dp[M][4][M];
long long compo(long long x, long long y) {
  if (x == y) return 0;
  if (x == 0) return 1;
  if (x == 3) return 1;
  if (x == 1 && y == 2) return 2;
  if (y == 1 && x == 2) return 2;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  {
    long long n, k;
    cin >> n >> k;
    for (long long j = 0; j < 4; j++) {
      if (j == 0 || j == 3)
        dp[1][j][1] += 1;
      else
        dp[1][j][2] += 1;
    }
    for (long long i = 2; i <= n; ++i) {
      for (long long j = 0; j < 4; j++) {
        for (long long k = 0; k < 4; k++) {
          for (long long l = 1; l <= 2 * n; l++) {
            long long x = l + compo(k, j);
            if (x > 2 * n) continue;
            dp[i][j][x] = (dp[i][j][x] + dp[i - 1][k][l]) % mod;
          }
        }
      }
    }
    long long ans = 0;
    for (long long j = 0; j < 4; j++) {
      ans = (ans + dp[n][j][k]) % mod;
    }
    cout << ans << "\n";
  }
  return 0;
}
