#include <bits/stdc++.h>
using namespace std;
const long long N = 1001;
const long long M = 998244353;
long long mod(long long m, long long mod = M) {
  m %= mod;
  return (m + mod) % mod;
}
long long dp[N][4][2 * N];
long long n, k;
long long solve(long long i, long long pre, long long cnt) {
  if (i >= n) return (cnt == k);
  long long &tmp = dp[i][pre][cnt];
  if (tmp != -1) return tmp;
  tmp = 0;
  for (long long j = 0; j < 4; j++) {
    if (pre == 3 or pre == 0)
      tmp = mod(tmp + solve(i + 1, j, cnt + (j != pre)));
    else
      tmp = mod(tmp + solve(i + 1, j, cnt + 2 * ((pre ^ j) == 3)));
  }
  return tmp;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  long long ans = 0;
  for (long long i = 0; i < 4; i++)
    ans = mod(ans + solve(1, i, 1 + (i == 1 or i == 2)));
  cout << ans << '\n';
}

### Prompt

Create a solution in CPP for 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;
const long long N = 1001;
const long long M = 998244353;
long long mod(long long m, long long mod = M) {
  m %= mod;
  return (m + mod) % mod;
}
long long dp[N][4][2 * N];
long long n, k;
long long solve(long long i, long long pre, long long cnt) {
  if (i >= n) return (cnt == k);
  long long &tmp = dp[i][pre][cnt];
  if (tmp != -1) return tmp;
  tmp = 0;
  for (long long j = 0; j < 4; j++) {
    if (pre == 3 or pre == 0)
      tmp = mod(tmp + solve(i + 1, j, cnt + (j != pre)));
    else
      tmp = mod(tmp + solve(i + 1, j, cnt + 2 * ((pre ^ j) == 3)));
  }
  return tmp;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  long long ans = 0;
  for (long long i = 0; i < 4; i++)
    ans = mod(ans + solve(1, i, 1 + (i == 1 or i == 2)));
  cout << ans << '\n';
}

```