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;
int k;
long long dp[2][2003][4];
int chk(int a, int b) {
  if (a == 0) return b == 3;
  if (a == 3) return b == 0;
  return __builtin_popcount(a ^ b);
}
int main() {
  int n;
  cin >> n >> k;
  dp[0][0][0] = dp[0][0][1] = dp[0][0][2] = dp[0][0][3] = 1;
  for (int ctr1 = 1; ctr1 < n; ++ctr1) {
    memset(dp[ctr1 & 1], 0, sizeof(dp[ctr1 & 1]));
    for (int ctr2 = 0; ctr2 <= k; ++ctr2) {
      for (int ctr3 = 0; ctr3 < 4; ++ctr3) {
        for (int ctr4 = 0; ctr4 < 4; ++ctr4) {
          dp[(ctr1 & 1)][ctr2 + chk(ctr3, ctr4)][ctr4] =
              (dp[(ctr1 & 1)][ctr2 + chk(ctr3, ctr4)][ctr4] +
               dp[!(ctr1 & 1)][ctr2][ctr3]) %
              998244353;
        }
      }
    }
  }
  long long rez = (dp[!(n & 1)][k - 1][0] + dp[!(n & 1)][k - 1][3]) % 998244353;
  if (k > 1)
    rez = (rez + dp[!(n & 1)][k - 2][1] + dp[!(n & 1)][k - 2][2]) % 998244353;
  cout << rez;
  return 0;
}
