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 <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 10005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353;
int N,X;
int dp[3005][6005],fac[6005],invfac[6005];
int f[3005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void Solve() {
    read(N);read(X);
    dp[0][0] = 1;
    fac[0] = 1;
    for(int i = 1 ; i <= 2 * N ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[2 * N] = fpow(fac[2 * N],MOD - 2);
    for(int i = 2 * N - 1 ; i >= 0 ; --i) {
	invfac[i] = mul(invfac[i + 1],i + 1);
    }
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
	for(int j = 2 * i ; j >= 1 ; --j) {
	    if(j >= 2) update(dp[i][j],dp[i - 1][j - 2]);
	    if(j >= 1) update(dp[i][j],dp[i - 1][j - 1]);
	}
    }
    for(int i = 0 ; i <= N ; ++i) {
	for(int j = 0 ; j <= 2 * i ; ++j) {
	    if(j < X) update(ans,mul(dp[i][j],C(N,i)));
	}
    }
    for(int i = 1 ; i < N ; ++i) {
	int t = i * 2 + 1;
	if(t >= X) break;
	for(int j = 0 ; j <= N ; ++j) {
	    if(i + j + 2 <= N) {
		update(f[i + j + 2],dp[j][X - 1 - t]);
		if(i + j + i + 2 <= N) update(f[i + j + i + 2],MOD - dp[j][X - 1 - t]);
	    }
	}
    }
    for(int i = 1 ; i <= N ; ++i) f[i] = inc(f[i],f[i - 1]);
    for(int i = 1 ; i <= N ; ++i) {
	update(ans,mul(f[i],C(N,i)));
    }
    if(X & 1) {
	for(int i = 1 ; i <= N ; ++i) {
	    if(i * 2 > X) update(ans,C(N,i));
	}
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
