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<iomanip>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<unordered_map>
#include<stack>
#include<queue>
#include<math.h>
#include<functional>
#include<bitset>
#include<cassert>
#include<random>
using namespace std;
using lint = long long;
using ld = long double;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
#define MOD 998244353LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(lint i=n;i<(int)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

#include <cstdint>
struct mint {
	using i64 = std::int_fast64_t;
	const static i64 mod = 998244353;
	i64 n;
public:
	mint(const i64 n = 0) : n((n % mod + mod) % mod) {}
	mint pow(int m) const {
		i64 a = n, r = 1;
		while(m > 0) {
			if(m & 1) { r *= a; r %= mod; }
			a = (a * a) % mod; m /= 2;
		}
		return mint(r);
	}
	mint &operator++() { *this += 1; return *this; }
	mint &operator--() { *this -= 1; return *this; }
	mint operator++(int) { mint ret = *this; *this += 1; return ret; }
	mint operator--(int) { mint ret = *this; *this -= 1; return ret; }
	mint operator~() const { return (this -> pow(mod - 2)); } // inverse
	friend bool operator==(const mint& lhs, const mint& rhs) {
		return lhs.n == rhs.n;
	}
	friend mint &operator+=(mint& lhs, const mint& rhs) {
		lhs.n += rhs.n;
		if (lhs.n >= mod) lhs.n -= mod;
		return lhs;
	}
	friend mint &operator-=(mint& lhs, const mint& rhs) {
		lhs.n -= rhs.n;
		if (lhs.n < 0) lhs.n += mod;
		return lhs;
	}
	friend mint &operator*=(mint& lhs, const mint& rhs) {
		lhs.n = (lhs.n * rhs.n) % mod;
		return lhs;
	}
	friend mint &operator/=(mint& lhs, const mint& rhs) {
		lhs.n = (lhs.n * (~rhs).n) % mod;
		return lhs;
	}
	friend mint operator+(const mint& lhs, const mint& rhs) {
		return mint(lhs.n + rhs.n);
	}
	friend mint operator-(const mint& lhs, const mint& rhs) {
		return mint(lhs.n - rhs.n);
	}
	friend mint operator*(const mint& lhs, const mint& rhs) {
		return mint(lhs.n * rhs.n);
	}
	friend mint operator/(const mint& lhs, const mint& rhs) {
		return mint(lhs.n * (~rhs).n);
	}
};
istream& operator>>(istream& is, mint& m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, mint& m) { os << m.n; return os; }

#define MAX_N 1000000
long long extgcd(long long a, long long b, long long& x, long long& y) {
	long long d = a;
	if (b != 0) {
		d = extgcd(b, a % b, y, x);
		y -= (a / b) * x;
	} else {
		x = 1; y = 0;
	}
	return d;
}
long long mod_inverse(long long a, long long m) {
	long long x, y;
	if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
	else return -1;
}
vector<long long> fact(MAX_N+1, INF);
long long mod_fact(long long n, long long& e) {
	if(fact[0] == INF) {
		fact[0]=1;
		if(MAX_N != 0) fact[1]=1;
		for(lint i = 2; i <= MAX_N; ++i) {
			fact[i] = (fact[i-1] * i) % MOD;
		}
	}
	e = 0;
	if(n == 0) return 1;
	long long res = mod_fact(n / MOD, e);
	e += n / MOD;
	if((n / MOD) % 2 != 0) return (res * (MOD - fact[n % MOD])) % MOD;
	return (res * fact[n % MOD]) % MOD;
}
// return nCk
long long mod_comb(long long n, long long k) {
	if(n < 0 || k < 0 || n < k) return 0;
	long long e1, e2, e3;
	long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
	if(e1 > e2 + e3) return 0;
	return (a1 * mod_inverse((a2 * a3) % MOD, MOD)) % MOD;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, x;
	cin >> n >> x;
	if(x == 2) {
		cout << 1 + n << endl;
		return 0;
	}
	mint ans = 1;
	FOR(m, 1, n + 1) {
		if(x == 1) {
			ans += mod_comb(n, m);
			continue;
		}
		mint tmp = 0;
		REP(i, m + 1) {
			if(i * 2 + (m - i) * 1 >= x) break;
			tmp += mod_comb(m, i);
		}
		FOR(i, 1, m) {
			if(i < m - i) {
				if(2 * i == x - 1) tmp++;
			} else {
				tmp += mod_comb(m - 2 * (m - i), x - 1 - 2 * (m - i) - (i - (m - i)));
			}
		}
		ans += tmp * mod_comb(n, m);
	}
	cout << ans << endl;
	return 0;
}
/* --------------------------------------- */
