#include <bits/stdc++.h>
using namespace std;
const int N = 2020, M = 4, MOD = 998244353;
int n, k, dp[N][N][M];
int sum(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
void _sum(int &a, int b) { a = sum(a, b); }
int check(int a, int b) {
  if (a == b) return 0;
  if (a == 3 || a == 2) a ^= 3, b ^= 3;
  if (a == 0) return 1;
  if (b == 2) return 2;
  return 0;
}
int main() {
  cin >> n >> k;
  dp[0][1][0] = dp[0][1][3] = 1;
  dp[0][2][1] = dp[0][2][2] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= k; j++)
      for (int a = 0; a < M; a++)
        for (int b = 0; b < M; b++)
          _sum(dp[i + 1][j + check(a, b)][b], dp[i][j][a]);
  int ans = 0;
  for (int i = 0; i < M; i++) _sum(ans, dp[n - 1][k][i]);
  cout << ans;
  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;
const int N = 2020, M = 4, MOD = 998244353;
int n, k, dp[N][N][M];
int sum(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
void _sum(int &a, int b) { a = sum(a, b); }
int check(int a, int b) {
  if (a == b) return 0;
  if (a == 3 || a == 2) a ^= 3, b ^= 3;
  if (a == 0) return 1;
  if (b == 2) return 2;
  return 0;
}
int main() {
  cin >> n >> k;
  dp[0][1][0] = dp[0][1][3] = 1;
  dp[0][2][1] = dp[0][2][2] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= k; j++)
      for (int a = 0; a < M; a++)
        for (int b = 0; b < M; b++)
          _sum(dp[i + 1][j + check(a, b)][b], dp[i][j][a]);
  int ans = 0;
  for (int i = 0; i < M; i++) _sum(ans, dp[n - 1][k][i]);
  cout << ans;
  return 0;
}

```