#include <bits/stdc++.h>
using namespace std;
long long dp[2001][2001][5];
int n, k;
long long MOD = 998244353;
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
int check(int a, int b) {
  int cnt = __builtin_popcount(a ^ b);
  if (cnt == 0) return cnt;
  if (cnt == 2)
    if (a == 0 || a == 3)
      return 1;
    else
      return 2;
  if (b == 0 || b == 3)
    return 0;
  else
    return 1;
}
long long f(int pos, int count, int bitmask) {
  long long ans = 0;
  if (pos == n) {
    if (count == k) return 1;
    return 0;
  }
  if (dp[pos][count][bitmask] != -1) return dp[pos][count][bitmask];
  if (pos == 0) {
    for (int i = 0; i < 4; i++) {
      if (i == 0 || i == 3)
        ans = add(ans, f(pos + 1, count + 1, i));
      else
        ans = add(ans, f(pos + 1, count + 2, i));
    }
    return dp[pos][count][bitmask] = ans;
  }
  for (int i = 0; i < 4; i++)
    ans = add(ans, f(pos + 1, count + check(bitmask, i), i));
  return dp[pos][count][bitmask] = ans;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  long long ans = 0;
  cout << f(0, 0, 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;
long long dp[2001][2001][5];
int n, k;
long long MOD = 998244353;
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
int check(int a, int b) {
  int cnt = __builtin_popcount(a ^ b);
  if (cnt == 0) return cnt;
  if (cnt == 2)
    if (a == 0 || a == 3)
      return 1;
    else
      return 2;
  if (b == 0 || b == 3)
    return 0;
  else
    return 1;
}
long long f(int pos, int count, int bitmask) {
  long long ans = 0;
  if (pos == n) {
    if (count == k) return 1;
    return 0;
  }
  if (dp[pos][count][bitmask] != -1) return dp[pos][count][bitmask];
  if (pos == 0) {
    for (int i = 0; i < 4; i++) {
      if (i == 0 || i == 3)
        ans = add(ans, f(pos + 1, count + 1, i));
      else
        ans = add(ans, f(pos + 1, count + 2, i));
    }
    return dp[pos][count][bitmask] = ans;
  }
  for (int i = 0; i < 4; i++)
    ans = add(ans, f(pos + 1, count + check(bitmask, i), i));
  return dp[pos][count][bitmask] = ans;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  long long ans = 0;
  cout << f(0, 0, 0);
}

```