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 dp[1002][2006][6], n, k, MOD = 998244353;
long long solve(int pos, int comp, int last) {
  if (pos == n) {
    return comp == k;
  }
  if (comp > k) {
    return 0;
  }
  if (dp[pos][comp][last] < 0) {
    dp[pos][comp][last] = 0;
    if (last == 0) {
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 0)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 1)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 2)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 3)) % MOD;
    }
    if (last == 1) {
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 0)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 1)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 2, 2)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 3)) % MOD;
    }
    if (last == 2) {
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 0)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 2, 1)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 2)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 3)) % MOD;
    }
    if (last == 3) {
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 0)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 1)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 2)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp, 3)) % MOD;
    }
    if (last == 4) {
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 0)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 2, 1)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 2, 2)) % MOD;
      dp[pos][comp][last] =
          (dp[pos][comp][last] + solve(pos + 1, comp + 1, 3)) % MOD;
    }
  }
  return dp[pos][comp][last];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  cout << solve(0, 0, 4) << endl;
}
