You are given a grid, consisting of 22 rows and nn 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 AA and BB belong to the same component if they are neighbours, or if there is a neighbour of AA that belongs to the same component with BB.

Let's call some bicoloring beautiful if it has exactly kk components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo 998244353998244353.

Input

The only line contains two integers nn and kk (1≤n≤10001≤n≤1000, 1≤k≤2n1≤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 998244353998244353.

Examples

Input

3 4

Output

12

Input

4 1

Output

2

Input

1 2

Output

2

題意：

有2行，n列，求k個連通塊的方法。

自己想實在是想不出來啊，看了看題解才勉強明白。

設一個三維dp陣列。

dp[i][j][k]表示前i列有j個聯通塊，k表示i列地板的顏色，k=0表示11,k=1表示10,k=2表示01,k=3表示00。

然後有下面的遞推關係：

 if(m==0)
       dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
 if(m==1)
       dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j-2][2]+dp[i-1][j-1][3])%mod;
 if(m==2)
       dp[i][j][2]=(dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;

if(m==3)
       dp[i][j][3]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j][3])%mod;

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const long long int mod=998244353;
const int maxn=1005;
int n,k;
ll dp[maxn][maxn<<1][4];
void init ()
{
    memset (dp,0,sizeof(dp));
    dp[1][1][0]=1;
    dp[1][2][1]=1;
    dp[1][2][2]=1;
    dp[1][1][3]=1;
}
int main()
{
    scanf("%d%d",&n,&k);
    init();
    for (int i=2;i<=n;i++)
    {
        for (int j=1;j<=k;j++)
        {
            for (int m=0;m<4;m++)
            {
                if(m==0)
                {
                    dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
                }
                else if(m==1)
                {
                    dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j-2][2]+dp[i-1][j-1][3])%mod;
                }
                else if(m==2)
                {
                    dp[i][j][2]=(dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
                }
                else
                {
                    dp[i][j][3]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j][3])%mod;
                }
            }
        }
    }
    printf("%lld\n",(dp[n][k][0]+dp[n][k][1]+dp[n][k][2]+dp[n][k][3])%mod);
    return 0;
}