#include <bits/stdc++.h>
using namespace std;
int dp[1005][2005][4];
long long MOD = 998244353;
int add(int second_last, int last) {
  switch (second_last) {
    case 0:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 1;
        case 2:
          return 1;
        case 3:
          return 1;
      }
    case 1:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 0;
        case 2:
          return 2;
        case 3:
          return 0;
      }
    case 2:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 2;
        case 2:
          return 0;
        case 3:
          return 0;
      }
    case 3:
      switch (last) {
        case 0:
          return 1;
        case 1:
          return 1;
        case 2:
          return 1;
        case 3:
          return 0;
      }
  }
}
int main() {
  long long n, k;
  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 column = 2; column <= n; column++) {
    for (int compo = 1; compo <= k; compo++) {
      for (int second_last = 0; second_last <= 3; second_last++) {
        for (int last = 0; last <= 3; last++) {
          long long new_compo = compo + add(second_last, last);
          dp[column][new_compo][last] += dp[column - 1][compo][second_last];
          dp[column][new_compo][last] %= MOD;
        }
      }
    }
  }
  long long ans = 0;
  for (int last = 0; last <= 3; last++) {
    ans += dp[n][k][last];
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a CPP solution 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>
using namespace std;
int dp[1005][2005][4];
long long MOD = 998244353;
int add(int second_last, int last) {
  switch (second_last) {
    case 0:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 1;
        case 2:
          return 1;
        case 3:
          return 1;
      }
    case 1:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 0;
        case 2:
          return 2;
        case 3:
          return 0;
      }
    case 2:
      switch (last) {
        case 0:
          return 0;
        case 1:
          return 2;
        case 2:
          return 0;
        case 3:
          return 0;
      }
    case 3:
      switch (last) {
        case 0:
          return 1;
        case 1:
          return 1;
        case 2:
          return 1;
        case 3:
          return 0;
      }
  }
}
int main() {
  long long n, k;
  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 column = 2; column <= n; column++) {
    for (int compo = 1; compo <= k; compo++) {
      for (int second_last = 0; second_last <= 3; second_last++) {
        for (int last = 0; last <= 3; last++) {
          long long new_compo = compo + add(second_last, last);
          dp[column][new_compo][last] += dp[column - 1][compo][second_last];
          dp[column][new_compo][last] %= MOD;
        }
      }
    }
  }
  long long ans = 0;
  for (int last = 0; last <= 3; last++) {
    ans += dp[n][k][last];
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}

```