#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
string s;
long long ans = 1, sim = 1, dif = 0, n, k, dp[1008][2008][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  dp[1][1][sim] = 2;
  dp[1][1][dif] = 0;
  dp[1][2][sim] = 0;
  dp[1][2][dif] = 2;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      dp[i][j][sim] += (dp[i - 1][j][sim] + 2 * dp[i - 1][j][dif]) % mod;
      dp[i][j][dif] += (dp[i - 1][j][dif]) % mod;
      dp[i][j + 1][dif] += (2 * dp[i - 1][j][sim]) % mod;
      dp[i][j + 1][sim] += (dp[i - 1][j][sim]) % mod;
      dp[i][j + 2][dif] += (dp[i - 1][j][dif]) % mod;
    }
  }
  cout << (dp[n][k][sim] + dp[n][k][dif]) % mod << '\n';
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
long long mod = 998244353;
string s;
long long ans = 1, sim = 1, dif = 0, n, k, dp[1008][2008][4];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  dp[1][1][sim] = 2;
  dp[1][1][dif] = 0;
  dp[1][2][sim] = 0;
  dp[1][2][dif] = 2;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      dp[i][j][sim] += (dp[i - 1][j][sim] + 2 * dp[i - 1][j][dif]) % mod;
      dp[i][j][dif] += (dp[i - 1][j][dif]) % mod;
      dp[i][j + 1][dif] += (2 * dp[i - 1][j][sim]) % mod;
      dp[i][j + 1][sim] += (dp[i - 1][j][sim]) % mod;
      dp[i][j + 2][dif] += (dp[i - 1][j][dif]) % mod;
    }
  }
  cout << (dp[n][k][sim] + dp[n][k][dif]) % mod << '\n';
  return 0;
}

```