Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.

Constraints

* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.

Input

Input is given from Standard Input in the following format:


N X


Output

Print the number, modulo 998244353, of sequences that satisfy the condition.

Examples

Input

3 3


Output

14


Input

8 6


Output

1179


Input

10 1


Output

1024


Input

9 13


Output

18402


Input

314 159


Output

459765451
#include <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <utility>
#include <queue>
#include <set>
#include <map>
#include <deque>
#include <iomanip>
#include <cstdio>
#include <stack>
#include <numeric>

using namespace std;
typedef  long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define  MP make_pair
#define  PB push_back
#define inf 1000000007
#define mod 998244353
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define MAX_N 100005
#define MOD 998244353

int inv[MAX_N],fac[MAX_N],finv[MAX_N];

void make()
{
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for(int i=2;i<MAX_N;i++){
		inv[i] = MOD - (long long)inv[MOD%i] * (MOD/i) % MOD;
		fac[i] = (long long)fac[i-1] * i % MOD;
		finv[i] = (long long)finv[i-1] * inv[i] % MOD;
	}
}

int comb(int a,int b)
{
	if(a < b){
		return 0;
	}
	return fac[a] * ((long long)finv[b] * finv[a-b] % MOD) % MOD;
}
ll dp[3010][6020];
int main(){
    make();
    int n;
    int x;
    cin >> n >> x;
    dp[0][0] = 1;
    for(int i=0;i<n;i++){
        for(int j=0;j<=2*i;j++){
            dp[i+1][j+1] += dp[i][j];
            dp[i+1][j+1] %= mod;
            dp[i+1][j+2] += dp[i][j];
            dp[i+1][j+2] %= mod;
        }
    }
    ll ans = 0;
    for(int i=0;i<=n;i++){
        for(int j=0;j<x;j++){
            ans += dp[i][j]*comb(n,i)%mod;
            ans %= mod;
        }
    }
    cerr << ans << endl;
    for(int k=1;k<=(n)/2;k++){
        for(int p = 0;p<=n-2*k;p++){
            if(x-1-2*k<=0)continue;
            ans += dp[p][x-1-2*k]*comb(n,p+2*k)%mod;
            ans %= mod;
            if(2*p==x-1-2*k){
                ans += mod-comb(n,p+2*k);
                ans %= mod;
            }
        }
        
    }
    cerr << ans << endl;
    if(x%2==0){
        
    }else{
        for(int k = (x+1)/2;k<=n;k++){
            
            ans += comb(n,k);
            ans %= mod;
        }
    }
    cout << ans << endl;
    return 0;
}