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 dp1[2 * 1001][4], dp2[2 * 1001][4];
int num[4][4];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  if (n == 1) {
    if (k == 1)
      cout << 2 << endl;
    else
      cout << 2 << endl;
    return 0;
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      int i0 = i % 2, i1 = i / 2;
      int j0 = j % 2, j1 = j / 2;
      if (j0 == j1 && (i0 == j0 || i1 == j1))
        num[i][j] = 0;
      else if (j0 == j1)
        num[i][j] = 1;
      else {
        if (i0 != j0) num[i][j] += 1;
        if (i1 != j1) num[i][j] += 1;
      }
      if (i == 0 || i == 3)
        dp2[1 + num[i][j]][j] += 1;
      else
        dp2[2 + num[i][j]][j] += 1;
    }
  }
  for (int a = 3; a <= n; a++) {
    for (int b = 0; b <= k; b++) {
      for (int j = 0; j < 4; j++) {
        for (int i = 0; i < 4; i++) {
          if (b >= num[i][j]) {
            dp1[b][j] += dp2[b - num[i][j]][i];
            if (dp1[b][j] >= 998244353LL) dp1[b][j] -= 998244353LL;
          }
        }
      }
    }
    for (int b = 0; b <= k; b++) {
      for (int j = 0; j < 4; j++) {
        dp2[b][j] = dp1[b][j];
        dp1[b][j] = 0;
      }
    }
  }
  cout << (dp2[k][0] + dp2[k][1] + dp2[k][2] + dp2[k][3]) % 998244353LL << endl;
}
