#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, k;
long long dp[N][2 * N][3][3];
long long md(long long x) { return x % 998244353LL; }
long long solve(int p, int x, int flag1, int flag2) {
  if (x > k) return 0LL;
  if (p == n) {
    if (x == k) return 1LL;
    return 0LL;
  }
  long long &ans = dp[p][x][flag1][flag2];
  if (ans != -1LL) return ans;
  int at = 0;
  ans = 0LL;
  if (flag1 != 1 && flag2 != 1) at = 1;
  ans = md(ans + solve(p + 1, x + at, 1, 1));
  at = 0;
  if (flag1 != 0 && flag2 != 0) at = 1;
  ans = md(ans + solve(p + 1, x + at, 0, 0));
  at = 0;
  if (flag1 != 1) at++;
  if (flag2 != 0) at++;
  ans = md(ans + solve(p + 1, x + at, 1, 0));
  at = 0;
  if (flag1 != 0) at++;
  if (flag2 != 1) at++;
  ans = md(ans + solve(p + 1, x + at, 0, 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  memset(dp, -1LL, sizeof(dp));
  cin >> n >> k;
  cout << solve(0, 0, 2, 2) << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int N = 1010;
int n, k;
long long dp[N][2 * N][3][3];
long long md(long long x) { return x % 998244353LL; }
long long solve(int p, int x, int flag1, int flag2) {
  if (x > k) return 0LL;
  if (p == n) {
    if (x == k) return 1LL;
    return 0LL;
  }
  long long &ans = dp[p][x][flag1][flag2];
  if (ans != -1LL) return ans;
  int at = 0;
  ans = 0LL;
  if (flag1 != 1 && flag2 != 1) at = 1;
  ans = md(ans + solve(p + 1, x + at, 1, 1));
  at = 0;
  if (flag1 != 0 && flag2 != 0) at = 1;
  ans = md(ans + solve(p + 1, x + at, 0, 0));
  at = 0;
  if (flag1 != 1) at++;
  if (flag2 != 0) at++;
  ans = md(ans + solve(p + 1, x + at, 1, 0));
  at = 0;
  if (flag1 != 0) at++;
  if (flag2 != 1) at++;
  ans = md(ans + solve(p + 1, x + at, 0, 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  memset(dp, -1LL, sizeof(dp));
  cin >> n >> k;
  cout << solve(0, 0, 2, 2) << "\n";
  return 0;
}

```