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 MOD = 998244353;
const int N = 2018;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int dp[N][N][4];
int more[4][4];
int n, k;
int main() {
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < 4; ++j) {
      if ((j & 1) != (j >> 1)) {
        if ((j & 1) != (i & 1)) more[i][j]++;
        if ((j >> 1) != (i >> 1)) more[i][j]++;
      } else {
        if ((i & 1) == (i >> 1) && (i & 1) != (j & 1)) more[i][j]++;
      }
    }
  }
  cin >> n >> k;
  for (int mask = 0; mask < 4; ++mask) {
    dp[1][1 + ((mask & 1) != (mask >> 1))][mask]++;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 2 * i; ++j) {
      for (int mask = 0; mask < 4; ++mask) {
        if (dp[i][j][mask]) {
          for (int new_mask = 0; new_mask < 4; ++new_mask) {
            add(dp[i + 1][j + more[mask][new_mask]][new_mask], dp[i][j][mask]);
          }
        }
      }
    }
  }
  int ans = 0;
  for (int mask = 0; mask < 4; ++mask) {
    add(ans, dp[n][k][mask]);
  }
  cout << ans << endl;
  return 0;
}
