#include <bits/stdc++.h>
int dp[1003][2006][4] = {0};
int main() {
  int n, lm;
  scanf("%d %d", &n, &lm);
  dp[1][1][0] = 1;
  dp[1][2][2] = 1;
  dp[1][2][1] = 1;
  dp[1][1][3] = 1;
  long long temp = 0;
  for (int i = 2; i <= n; ++i) {
    for (int k = 1; k <= (i << 1); ++k) {
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k][0] + dp[i - 1][k][2] +
             dp[i - 1][k - 1][3];
      dp[i][k][0] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k - 1][0] +
             dp[i - 1][k - 2][2] + dp[i - 1][k - 1][3];
      dp[i][k][1] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k - 2][1] + dp[i - 1][k - 1][0] +
             dp[i - 1][k][2] + dp[i - 1][k - 1][3];
      dp[i][k][2] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k - 1][0] + dp[i - 1][k][2] +
             dp[i - 1][k][3];
      dp[i][k][3] = temp % 998244353;
      temp = 0;
    }
  }
  long long ans = 0;
  ans = ans + dp[n][lm][0] + dp[n][lm][1] + dp[n][lm][2] + dp[n][lm][3];
  ans = ans % 998244353;
  printf("%I64d\n", ans);
}

### Prompt

Please create a solution in Cpp to 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>
int dp[1003][2006][4] = {0};
int main() {
  int n, lm;
  scanf("%d %d", &n, &lm);
  dp[1][1][0] = 1;
  dp[1][2][2] = 1;
  dp[1][2][1] = 1;
  dp[1][1][3] = 1;
  long long temp = 0;
  for (int i = 2; i <= n; ++i) {
    for (int k = 1; k <= (i << 1); ++k) {
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k][0] + dp[i - 1][k][2] +
             dp[i - 1][k - 1][3];
      dp[i][k][0] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k - 1][0] +
             dp[i - 1][k - 2][2] + dp[i - 1][k - 1][3];
      dp[i][k][1] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k - 2][1] + dp[i - 1][k - 1][0] +
             dp[i - 1][k][2] + dp[i - 1][k - 1][3];
      dp[i][k][2] = temp % 998244353;
      temp = 0;
      temp = temp + dp[i - 1][k][1] + dp[i - 1][k - 1][0] + dp[i - 1][k][2] +
             dp[i - 1][k][3];
      dp[i][k][3] = temp % 998244353;
      temp = 0;
    }
  }
  long long ans = 0;
  ans = ans + dp[n][lm][0] + dp[n][lm][1] + dp[n][lm][2] + dp[n][lm][3];
  ans = ans % 998244353;
  printf("%I64d\n", ans);
}

```