#include <bits/stdc++.h>
using namespace std;
vector<int> adj[100001];
bool visit[100001];
const int mod = 998244353;
long long mod_mult(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long mult(long long x, long long y) {
  long long prod = 1;
  while (y > 0) {
    if (y & 1) prod = mod_mult(prod, x);
    x = mod_mult(x, x);
    y /= 2;
  }
  return prod;
}
long long dp[1002][2002][4];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, i;
  cin >> n >> k;
  dp[0][1][0] = dp[0][1][3] = 1;
  dp[0][2][2] = dp[0][2][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][2] +
                    dp[i - 1][j][1];
      if (j - 2 >= 0) {
        dp[i][j][1] = dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                      dp[i - 1][j - 2][2] + dp[i - 1][j][1];
        dp[i][j][2] = dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                      dp[i - 1][j][2] + dp[i - 1][j - 2][1];
      } else {
        dp[i][j][1] =
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][1];
        dp[i][j][2] =
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][2];
      }
      dp[i][j][3] = dp[i - 1][j - 1][0] + dp[i - 1][j][3] + dp[i - 1][j][2] +
                    dp[i - 1][j][1];
      dp[i][j][0] %= mod;
      dp[i][j][1] %= mod;
      dp[i][j][2] %= mod;
      dp[i][j][3] %= mod;
    }
  }
  cout << (dp[n - 1][k][0] + dp[n - 1][k][1] + dp[n - 1][k][2] +
           dp[n - 1][k][3]) %
              mod;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
vector<int> adj[100001];
bool visit[100001];
const int mod = 998244353;
long long mod_mult(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long mult(long long x, long long y) {
  long long prod = 1;
  while (y > 0) {
    if (y & 1) prod = mod_mult(prod, x);
    x = mod_mult(x, x);
    y /= 2;
  }
  return prod;
}
long long dp[1002][2002][4];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, i;
  cin >> n >> k;
  dp[0][1][0] = dp[0][1][3] = 1;
  dp[0][2][2] = dp[0][2][1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][2] +
                    dp[i - 1][j][1];
      if (j - 2 >= 0) {
        dp[i][j][1] = dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                      dp[i - 1][j - 2][2] + dp[i - 1][j][1];
        dp[i][j][2] = dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] +
                      dp[i - 1][j][2] + dp[i - 1][j - 2][1];
      } else {
        dp[i][j][1] =
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][1];
        dp[i][j][2] =
            dp[i - 1][j - 1][0] + dp[i - 1][j - 1][3] + dp[i - 1][j][2];
      }
      dp[i][j][3] = dp[i - 1][j - 1][0] + dp[i - 1][j][3] + dp[i - 1][j][2] +
                    dp[i - 1][j][1];
      dp[i][j][0] %= mod;
      dp[i][j][1] %= mod;
      dp[i][j][2] %= mod;
      dp[i][j][3] %= mod;
    }
  }
  cout << (dp[n - 1][k][0] + dp[n - 1][k][1] + dp[n - 1][k][2] +
           dp[n - 1][k][3]) %
              mod;
  return 0;
}

```