#include <bits/stdc++.h>
const int Maxn = 100010;
const int p = 998244353;
using namespace std;
int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') x = x * 10 + s - '0', s = getchar();
  x *= f;
  return x;
}
void put(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) put(x / 10);
  putchar(x % 10 + '0');
}
long long dp[2][2001][4];
int n, k;
int main() {
  n = read();
  k = read();
  dp[1][1][0] = dp[1][1][3] = dp[1][2][1] = dp[1][2][2] = 1;
  int now = 0, pre = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[now][j][0] =
          (dp[pre][j][0] + dp[pre][j][1] + dp[pre][j][2] + dp[pre][j - 1][3]) %
          p;
      if (j > 1)
        dp[now][j][1] = (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j - 2][2] +
                         dp[pre][j - 1][3]) %
                        p,
        dp[now][j][2] = (dp[pre][j - 1][0] + dp[pre][j - 2][1] + dp[pre][j][2] +
                         dp[pre][j - 1][3]) %
                        p;
      else
        dp[now][j][1] =
            (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j - 1][3]) % p,
        dp[now][j][2] =
            (dp[pre][j - 1][0] + dp[pre][j][2] + dp[pre][j - 1][3]) % p;
      dp[now][j][3] =
          (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j][2] + dp[pre][j][3]) %
          p;
    }
    swap(now, pre);
  }
  put((dp[n & 1][k][0] + dp[n & 1][k][1] + dp[n & 1][k][2] + dp[n & 1][k][3]) %
      p);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int Maxn = 100010;
const int p = 998244353;
using namespace std;
int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') x = x * 10 + s - '0', s = getchar();
  x *= f;
  return x;
}
void put(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) put(x / 10);
  putchar(x % 10 + '0');
}
long long dp[2][2001][4];
int n, k;
int main() {
  n = read();
  k = read();
  dp[1][1][0] = dp[1][1][3] = dp[1][2][1] = dp[1][2][2] = 1;
  int now = 0, pre = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[now][j][0] =
          (dp[pre][j][0] + dp[pre][j][1] + dp[pre][j][2] + dp[pre][j - 1][3]) %
          p;
      if (j > 1)
        dp[now][j][1] = (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j - 2][2] +
                         dp[pre][j - 1][3]) %
                        p,
        dp[now][j][2] = (dp[pre][j - 1][0] + dp[pre][j - 2][1] + dp[pre][j][2] +
                         dp[pre][j - 1][3]) %
                        p;
      else
        dp[now][j][1] =
            (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j - 1][3]) % p,
        dp[now][j][2] =
            (dp[pre][j - 1][0] + dp[pre][j][2] + dp[pre][j - 1][3]) % p;
      dp[now][j][3] =
          (dp[pre][j - 1][0] + dp[pre][j][1] + dp[pre][j][2] + dp[pre][j][3]) %
          p;
    }
    swap(now, pre);
  }
  put((dp[n & 1][k][0] + dp[n & 1][k][1] + dp[n & 1][k][2] + dp[n & 1][k][3]) %
      p);
  return 0;
}

```