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 int N = 1e3;
int dp[N + 13][2 * N + 13][(1 << (2)) + 13];
const int base = 998244353;
int n, k;
int cntBit(int mask) { return __builtin_popcount(mask); }
int cal(int state, int mask) {
  if ((((mask) >> 0) & 1) == (((mask) >> 1) & 1)) {
    if (cntBit(mask ^ state) == 2)
      return 1;
    else
      return 0;
  } else
    return cntBit(mask ^ state);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  dp[1][1][0] = dp[1][2][1] = dp[1][2][2] = dp[1][1][3] = 1;
  for (int i = 1; i < n; ++i)
    for (int j = 1; j <= k; ++j)
      for (int mask = 0; mask < 4; ++mask) {
        if (i == 1 && j == 2 && mask == 1) cerr << 123;
        if (!dp[i][j][mask]) continue;
        for (int state = 0; state < 4; ++state) {
          int dist = cal(mask, state);
          if (j + dist > k) continue;
          dp[i + 1][j + dist][state] =
              (dp[i + 1][j + dist][state] + dp[i][j][mask]) % base;
        }
      }
  int res = 0;
  for (int mask = 0; mask < 4; ++mask) res = (res + dp[n][k][mask]) % base;
  cout << res % base;
}
