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 double PI = acos(-1.0);
void fast_in_out() {
  std::ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
}
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
int dy[] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
int lx[] = {2, 2, -1, 1, -2, -2, -1, 1};
int ly[] = {-1, 1, 2, 2, 1, -1, -2, -2};
const int N = 1e6 + 10, M = 1000 + 5, mod = 998244353;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int dp[2][M + M][4];
int main() {
  fast_in_out();
  int n, k;
  cin >> n >> k;
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  int m[4][4] = {
      {0, 0, 0, -1}, {-1, 0, -2, -1}, {-1, -2, 0, -1}, {-1, 0, 0, 0}};
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= (i + i); ++j) {
      int id1 = i & 1, id2 = id1 ^ 1;
      for (int mask1 = 0; mask1 < 4; mask1++) {
        dp[id1][j][mask1] = 0;
        for (int mask2 = 0; mask2 < 4; mask2++) {
          int nj = j + m[mask1][mask2];
          if (nj < 0) continue;
          dp[id1][j][mask1] += dp[id2][nj][mask2];
          if (dp[id1][j][mask1] >= mod) dp[id1][j][mask1] -= mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 4; i++) {
    ans += dp[n & 1][k][i];
    if (ans >= mod) ans -= mod;
  }
  cout << ans;
  return 0;
}
