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 mod = 998244353;
long long dp[1001][2001][4];
long long solve(long long n, long long k, long long i) {
  if (dp[n][k][i] != -1) {
    return dp[n][k][i];
  }
  if (k < 1 || k > 2 * n) return 0;
  if (i == 0) {
    dp[n][k][i] = solve(n - 1, k, 0) + solve(n - 1, k - 1, 1) +
                  solve(n - 1, k - 1, 2) + solve(n - 1, k - 2, 3);
  } else if (i == 1) {
    dp[n][k][i] = solve(n - 1, k, 0) + solve(n - 1, k, 1) +
                  solve(n - 1, k - 1, 2) + solve(n - 1, k, 3);
  } else if (i == 2) {
    dp[n][k][i] = solve(n - 1, k, 0) + solve(n - 1, k - 1, 1) +
                  solve(n - 1, k, 2) + solve(n - 1, k, 3);
  } else {
    dp[n][k][i] = solve(n - 1, k - 2, 0) + solve(n - 1, k - 1, 1) +
                  solve(n - 1, k - 1, 2) + solve(n - 1, k, 3);
  }
  dp[n][k][i] %= mod;
  return dp[n][k][i];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(dp, -1, sizeof dp);
  long long n, k;
  cin >> n >> k;
  dp[1][1][0] = dp[1][1][3] = 0;
  dp[1][1][1] = dp[1][1][2] = 1;
  dp[1][2][0] = dp[1][2][3] = 1;
  dp[1][2][1] = dp[1][2][2] = 0;
  cout << (solve(n, k, 0) + solve(n, k, 1) + solve(n, k, 2) + solve(n, k, 3)) %
              mod
       << endl;
}
