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;
int n;
int k;
long long m = 998244353;
long long bb[1001][3000];
long long bw[1001][3000];
long long wb[1001][3000];
long long ww[1001][3000];
long long sww(long long i, int c);
long long sbw(long long i, int c);
long long swb(long long i, int c);
long long sbb(long long i, int c) {
  if (i == n) {
    if (c == k) return 1;
    if (c + 1 == k) return 3;
    return 0;
  }
  if (bb[i][c] != -1) return bb[i][c];
  bb[i][c] =
      sbb(i + 1, c) + sww(i + 1, c + 1) + sbw(i + 1, c + 1) + swb(i + 1, c + 1);
  bb[i][c] %= m;
  return bb[i][c];
}
long long sww(long long i, int c) {
  if (i == n) {
    if (c == k) return 1;
    if (c + 1 == k) return 3;
    return 0;
  }
  if (ww[i][c] != -1) return ww[i][c];
  ww[i][c] =
      sbb(i + 1, c + 1) + sww(i + 1, c) + sbw(i + 1, c + 1) + swb(i + 1, c + 1);
  ww[i][c] %= m;
  return ww[i][c];
}
long long sbw(long long i, int c) {
  if (i == n) {
    if (c == k) return 3;
    if (c + 2 == k) return 1;
    return 0;
  }
  if (bw[i][c] != -1) return bw[i][c];
  bw[i][c] = sbb(i + 1, c) + sww(i + 1, c) + sbw(i + 1, c) + swb(i + 1, c + 2);
  bw[i][c] %= m;
  return bw[i][c];
}
long long swb(long long i, int c) {
  if (i == n) {
    if (c == k) return 3;
    if (c + 2 == k) return 1;
    return 0;
  }
  if (wb[i][c] != -1) return wb[i][c];
  wb[i][c] = sbb(i + 1, c) + sww(i + 1, c) + sbw(i + 1, c + 2) + swb(i + 1, c);
  wb[i][c] %= m;
  return wb[i][c];
}
int main() {
  cin >> n >> k;
  for (int i = 0; i <= 1000; i++) {
    for (int j = 0; j < 3000; j++) {
      bb[i][j] = -1;
      ww[i][j] = -1;
      bw[i][j] = -1;
      wb[i][j] = -1;
    }
  }
  if (n == 1) {
    if (k > 2) cout << 0 << endl;
    cout << 2 << endl;
    return 0;
  }
  cout << (sbw(2, 2) + swb(2, 2) + sww(2, 1) + sbb(2, 1)) % m << endl;
}
