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;
long long int mx(long long int a[], long long int n) {
  return *(max_element(a, a + n));
}
long long int mi(long long int a[], long long int n) {
  return *(min_element(a, a + n));
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % 998244353;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y >> 1;
    x = (x * x) % 998244353;
  }
  return res % 998244353;
}
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
long long int upperIndex(long long int arr[], long long int n,
                         long long int y) {
  long long int l = 0, h = n;
  while (l <= h) {
    long long int mid = (l + h) / 2;
    if (arr[mid] < y)
      l = mid + 1;
    else
      h = mid - 1;
  }
  return h;
}
void pa(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void ga(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
template <typename T>
void print_queue(T& q) {
  while (!q.empty()) {
    std::cout << q.top() << " ";
    q.pop();
  }
  std::cout << '\n';
}
long long int dp[1001][2001][4];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  dp[0][1][0] = 1;
  dp[0][2][1] = 1;
  dp[0][2][2] = 1;
  dp[0][1][3] = 1;
  for (long long int i = 1; i <= n - 1; i++) {
    for (long long int j = 1; j <= k; j++) {
      long long int pp = 0;
      dp[i][j][0] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j - 1][3]) %
                    998244353;
      dp[i][j][1] = (dp[i - 1][j - 1][0] + dp[i - 1][j][1] +
                     dp[i - 1][max(j - 2, pp)][2] + dp[i - 1][j - 1][3]) %
                    998244353;
      dp[i][j][2] = (dp[i - 1][j - 1][0] + dp[i - 1][max(j - 2, pp)][1] +
                     dp[i - 1][j][2] + dp[i - 1][j - 1][3]) %
                    998244353;
      dp[i][j][3] = (dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j][3]) %
                    998244353;
    }
  }
  cout << (dp[n - 1][k][0] + dp[n - 1][k][1] + dp[n - 1][k][2] +
           dp[n - 1][k][3]) %
              998244353;
}
