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

給一個2*n的矩陣，每個格子可以塗成黑或白色，問有k個聯通塊的塗色方式有幾種。

很容易想到dp，只需要記錄最後一列是黑黑、黑白、白黑、還是白白即可。

用0,1,2,3分別表示，則dp[i][j][k]表示在2*i的矩陣中，有j個聯通塊且,最後一行塗成k的方式。

dp即可

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define ull unsigend long long
const int M=1e3+5;
const int N=998244353;
ll dp[M][M*2][4],ans;
void init()
{
    ans=0;
    memset(dp,0,sizeof(dp));
}
ll solve(int n,int m)
{
    dp[1][1][0]=1;
    dp[1][1][3]=1;
    dp[1][2][1]=1;
    dp[1][2][2]=1;
    for(int i=2; i<=n; i++)
    {
        for(int j=1; j<=i*2; j++)
        {
            dp[i][j][0]+=dp[i-1][j][0];
            dp[i][j][0]+=dp[i-1][j][1];
            dp[i][j][0]+=dp[i-1][j][2];
            if(j>1) dp[i][j][0]+=dp[i-1][j-1][3];

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

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

            dp[i][j][3]+=dp[i-1][j][3];
            dp[i][j][3]+=dp[i-1][j][1];
            dp[i][j][3]+=dp[i-1][j][2];
            if(j>1) dp[i][j][3]+=dp[i-1][j-1][0];
            for(int k=0; k<3; k++) dp[i][j][k]%=N;
        }
    }
    for(int i=0; i<4; i++) ans+=dp[n][m][i];
    ans%=N;
    printf("%lld\n",ans);
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        solve(n,m);
    }
    return 0;
}