Codeforces 1051D Bicolorings

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

 

題意：給你一個2*n的矩陣，你只能填黑塊或者是白塊使得把這個矩陣分成k個部分，問你有多少種方案數？

 

 

解題思路：我設一個dp

dp[i][j][ki] 表示的是我到第i列已經有k塊了  ki==1的表示第i列不是同一種顏色，ki==0表示第i列是同一種顏色

由於第i行的狀態只與第i-1行的狀態有關m，所以我們只需要考慮2*2的矩陣的情況就行

首先 我們就對一列進行考慮，

很容易知道  dp[1][1][0]=2,dp[1][2][1]=2, 其他狀態都爲0

這時候我們考慮狀態轉移

2*2的矩陣中每一個格子可以選黑塊或者是白塊，所以有2*2*2*2=16種情況。

                                                   

    這四種爲同一種情況 dp[i][j][0]+=2*dp[i-1][j][1];

 

            

這兩種算一種情況  dp[i][j][0]+=dp[i-1][k-1][0];

         

這也算一種情況  dp[i][j][0]+=dp[i-1][j][0]; 

                                                                      

                              

同理 這也算一種情況

dp[i][j][1]+=2*dp[i-1][j-1][0];

 

                                    

這算一種情況，dp[i][j][1]+=dp[i-1][j][1];

 

                                       

最後一種情況

dp[i][j][1]+=dp[i-1][j-2][1];     

 

 

然後暴力dp求解

 

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=1100;
const long long mod=998244353;
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
ll dp[maxn][2*maxn][2],n,m;
int main(){
	int i,j;
	mem(dp,0);
	scanf("%I64d%I64d",&n,&m);
	dp[1][2][1]=2;dp[1][1][0]=2;
	for(i=2;i<=n;i++){
		for(j=1;j<=m;j++){
			dp[i][j][0]+=2*dp[i-1][j][1]+dp[i-1][j-1][0]+dp[i-1][j][0];
			dp[i][j][0]%=mod;
			dp[i][j][1]+=2*dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][1];
			dp[i][j][1]%=mod; 
		}
	}
	ll ans=(dp[n][m][0]+dp[n][m][1])%mod;
	printf("%I64d\n",ans);
	return 0;
}