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>
namespace chtholly {
inline int read() {
  int u = 0, dp = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) dp ^= c == '-';
  for (; isdigit(c); c = getchar()) u = (u << 3) + (u << 1) + (c ^ '0');
  return dp ? u : -u;
}
template <typename mitsuha>
inline bool read(mitsuha &u) {
  u = 0;
  int dp = 1;
  char c = getchar();
  for (; !isdigit(c) && ~c; c = getchar()) dp ^= c == '-';
  if (!~c) return 0;
  for (; isdigit(c); c = getchar()) u = (u << 3) + (u << 1) + (c ^ '0');
  return u = dp ? u : -u, 1;
}
template <typename mitsuha>
inline int write(mitsuha u) {
  if (!u) return 0 & putchar(48);
  if (u < 0) u = -u, putchar('-');
  int bit[20], i, p = 0;
  for (; u; u /= 10) bit[++p] = u % 10;
  for (i = p; i; --i) putchar(bit[i] + 48);
  return 0;
}
inline char fuhao() {
  char c = getchar();
  for (; isspace(c); c = getchar())
    ;
  return c;
}
}  // namespace chtholly
using namespace chtholly;
using namespace std;
const int aoi = 1018, mod = 998244353;
long long dp[aoi][aoi << 1][4];
int main() {
  int i, j, n = read(), k = read();
  dp[1][1][0] = dp[1][2][1] = dp[1][2][2] = dp[1][1][3] = 1;
  for (i = 2; i <= n; ++i) {
    for (j = 1; j <= 2 * i; ++j) {
      dp[i][j][0] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j - 1][3]) %
                    mod;
      dp[i][j][1] = (dp[i - 1][j - 1][0] + dp[i - 1][j][1] +
                     (j > 1 ? dp[i - 1][j - 2][2] : 0) + dp[i - 1][j - 1][3]) %
                    mod;
      dp[i][j][2] = (dp[i - 1][j - 1][0] + (j > 1 ? dp[i - 1][j - 2][1] : 0) +
                     dp[i - 1][j][2] + dp[i - 1][j - 1][3]) %
                    mod;
      dp[i][j][3] = (dp[i - 1][j - 1][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j][3]) %
                    mod;
    }
  }
  for (i = 0; i < 4; ++i) ***dp = (***dp + dp[n][k][i]) % mod;
  write(***dp);
}
