#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int MAX_N = 1005;
int dp[MAX_N][2 * MAX_N][4];
int n, k;
int cost[4][4];
int main() {
  cin >> n >> k;
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      cost[i][j] = (i != j);
    }
  }
  cost[2][0] = 0;
  cost[1][0] = 0;
  cost[1][3] = 0;
  cost[2][3] = 0;
  cost[1][2] = 2;
  cost[2][1] = 2;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      for (int x = 0; x < 4; x++) {
        if (dp[i][j][x] > 0) {
          for (int y = 0; y < 4; y++) {
            (dp[i + 1][j + cost[x][y]][y] += dp[i][j][x]) %= MOD;
          }
        }
      }
    }
  }
  int res = 0;
  for (int x = 0; x < 4; x++) {
    (res += dp[n][k][x]) %= MOD;
  }
  cout << res;
  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;
const int MOD = 998244353;
const int MAX_N = 1005;
int dp[MAX_N][2 * MAX_N][4];
int n, k;
int cost[4][4];
int main() {
  cin >> n >> k;
  dp[1][1][0] = 1;
  dp[1][2][1] = 1;
  dp[1][2][2] = 1;
  dp[1][1][3] = 1;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      cost[i][j] = (i != j);
    }
  }
  cost[2][0] = 0;
  cost[1][0] = 0;
  cost[1][3] = 0;
  cost[2][3] = 0;
  cost[1][2] = 2;
  cost[2][1] = 2;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      for (int x = 0; x < 4; x++) {
        if (dp[i][j][x] > 0) {
          for (int y = 0; y < 4; y++) {
            (dp[i + 1][j + cost[x][y]][y] += dp[i][j][x]) %= MOD;
          }
        }
      }
    }
  }
  int res = 0;
  for (int x = 0; x < 4; x++) {
    (res += dp[n][k][x]) %= MOD;
  }
  cout << res;
  return 0;
}

```