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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
inline int read() {
  int r = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    r = r * 10 + c - '0';
    c = getchar();
  }
  return r;
}
inline ll readll() {
  ll r = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    r = r * 10 + c - '0';
    c = getchar();
  }
  return r;
}
inline ll qpow(ll a, ll b, ll mod) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
inline ll gcd(ll a, ll b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
const double eps = 1e-8;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const ll mod = 998244353;
const int MAXN = 1e5;
const int MAXM = 1e5;
ll dp[1024][2048][2][2];
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  memset(dp, 0, sizeof(dp));
  dp[1][1][0][0] = 1;
  dp[1][1][1][1] = 1;
  dp[1][2][0][1] = 1;
  dp[1][2][1][0] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= 2 * i; ++j) {
      dp[i][j][0][0] = (dp[i - 1][j][0][0] + dp[i - 1][j][0][1] +
                        dp[i - 1][j][1][0] + dp[i - 1][j - 1][1][1]) %
                       mod;
      dp[i][j][1][1] = (dp[i - 1][j][1][1] + dp[i - 1][j][0][1] +
                        dp[i - 1][j][1][0] + dp[i - 1][j - 1][0][0]) %
                       mod;
      dp[i][j][1][0] = (dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][1] +
                        dp[i - 1][j][1][0] + dp[i - 1][j - 2][0][1]) %
                       mod;
      dp[i][j][0][1] = (dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][1] +
                        dp[i - 1][j][0][1] + dp[i - 1][j - 2][1][0]) %
                       mod;
    }
  }
  ll ans = 0;
  ans =
      (dp[n][k][0][0] + dp[n][k][1][0] + dp[n][k][0][1] + dp[n][k][1][1]) % mod;
  cout << ans << endl;
  return 0;
}
