#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
long long dp[1005][2005][4], n;
long long go(int id, int colors, int prev) {
  if (id == n or colors < 0) return colors == 0;
  if (dp[id][colors][prev] != -1) return dp[id][colors][prev];
  long long ans = 0;
  if (prev < 2) {
    for (int i = 0; i < 4; i++)
      ans += go(id + 1, colors - (i != prev), i) % MOD;
  } else {
    for (int i = 0; i < 4; i++)
      ans += go(id + 1, colors - 2 * (i > 1 && i != prev), i) % MOD;
  }
  return dp[id][colors][prev] = ans;
}
int main() {
  long long k, ans = 0;
  scanf("%lld %lld", &n, &k);
  memset(dp, -1, sizeof(dp));
  ans += go(1, k - 1, 0) % MOD;
  ans += go(1, k - 1, 1) % MOD;
  ans += go(1, k - 2, 2) % MOD;
  ans += go(1, k - 2, 3) % MOD;
  printf("%lld\n", ans % MOD);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
long long dp[1005][2005][4], n;
long long go(int id, int colors, int prev) {
  if (id == n or colors < 0) return colors == 0;
  if (dp[id][colors][prev] != -1) return dp[id][colors][prev];
  long long ans = 0;
  if (prev < 2) {
    for (int i = 0; i < 4; i++)
      ans += go(id + 1, colors - (i != prev), i) % MOD;
  } else {
    for (int i = 0; i < 4; i++)
      ans += go(id + 1, colors - 2 * (i > 1 && i != prev), i) % MOD;
  }
  return dp[id][colors][prev] = ans;
}
int main() {
  long long k, ans = 0;
  scanf("%lld %lld", &n, &k);
  memset(dp, -1, sizeof(dp));
  ans += go(1, k - 1, 0) % MOD;
  ans += go(1, k - 1, 1) % MOD;
  ans += go(1, k - 2, 2) % MOD;
  ans += go(1, k - 2, 3) % MOD;
  printf("%lld\n", ans % MOD);
  return 0;
}

```