CF1051D Bicolorings

原题链接：http://codeforces.com/contest/1051/problem/D

Bicolorings

You are given a grid, consisting of 222 rows and nnn 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 AAA and BBB belong to the same component if they are neighbours, or if there is a neighbour of AAA that belongs to the same component with BBB.

Let’s call some bicoloring beautiful if it has exactly kkk components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo 998244353998244353998244353.

Input

The only line contains two integers nnn and k(1≤n≤1000,1≤k≤2n)k (1≤n≤1000, 1≤k≤2n)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 998244353998244353998244353.

Examples

input

3 4

output

input

4 1

output

input

1 2

output

Note

One of possible bicolorings in sample 111:

题解

一眼状压dpdpdp题，dp[i][j][k]dp[i][j][k]dp[i][j][k]表示前iii列，联通块数量为jjj，该列的maskmaskmask为kkk时的方案数，日常转移没有什么问题.

代码

#include<bits/stdc++.h>
using namespace std;
const int M=1005,mod=998244353;
long long dp[M][M<<1][4];
int n,m,i,j,k;
void in(){scanf("%d%d",&n,&m);}
void ac()
{dp[1][1][0]=dp[1][1][3]=dp[1][2][1]=dp[1][2][2]=1;for(i=1;i<n;++i)for(j=1;j<=(i<<1);++j){for(k=0;k<4;++k)(dp[i+1][j][k]+=dp[i][j][k])%=mod;for(k=1;k<4;++k)(dp[i+1][j+1][k]+=dp[i][j][0])%=mod;for(k=1;k<3;++k)(dp[i+1][j+1][k]+=dp[i][j][3])%=mod,(dp[i+1][j][3]+=dp[i][j][k])%=mod,(dp[i+1][j][0]+=dp[i][j][k])%=mod;(dp[i+1][j+2][1]+=dp[i][j][2])%=mod,(dp[i+1][j+2][2]+=dp[i][j][1])%=mod,(dp[i+1][j+1][0]+=dp[i][j][3]);}printf("%lld",(dp[n][m][0]+dp[n][m][1]+dp[n][m][2]+dp[n][m][3])%mod);
}
int main(){in();ac();}