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
#define y1 askjdkasldjlkasd
#include <bits/stdc++.h>
#undef y1

using namespace std;

#define pb push_back
#define mp make_pair
#define fi(a, b) for(int i=a; i<=b; i++)
#define fj(a, b) for(int j=a; j<=b; j++)
#define fo(a, b) for(int o=a; o<=b; o++)
#define fdi(a, b) for(int i=a; i>=b; i--)
#define fdj(a, b) for(int j=a; j>=b; j--)
#define fdo(a, b) for(int o=a; o>=b; o--)
#define sz(x) (int)x.size()

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef vector<ll> vl;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<ll> vll;

#ifdef LOCAL
#define err(...) fprintf(stderr, __VA_ARGS__)
#else
#define err(...) while (0)
#endif

double START_TIME;

void exit() {
#ifdef LOCAL	
	cerr << "TIME: " << setprecision(5) << fixed << (clock() - START_TIME) / CLOCKS_PER_SEC << endl;
#endif	
	exit(0);
}

template<typename A, typename B>
ostream& operator<<(ostream& os, pair<A, B> p) {
	os << "(" << p.first << ", " << p.second << ")";
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, vector<T> v) {
	fi(0, sz(v) - 1) {
		os << v[i] << " ";
	}
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, set<T> t) {
	for (auto z : t) {
		os << z << " ";
	}
	return os;
}

template<typename T1, typename T2>
ostream& operator<<(ostream& os, map<T1, T2> t) {
	cerr << endl;
	for (auto z : t) {
		os << "\t" << z.first << " -> " << z.second << endl;
	}
	return os;
}

#ifdef LOCAL
#define dbg(x) {cerr << __LINE__ << "\t" << #x << ": " << x << endl;}
#define dbg0(x, n) {cerr << __LINE__ << "\t" << #x << ": "; for (int ABC = 0; ABC < n; ABC++) cerr << x[ABC] << ' '; cerr << endl;}
#else
#define dbg(x) while(0){}
#define dbg0(x, n) while(0){}
#endif

#ifdef LOCAL
#define ass(x) if (!(x)) { cerr << __LINE__ << "\tassertion failed: " << #x << endl, abort(); }
#else
#define ass(x) assert(x)
#endif

///////////////////////////////////////////////////

const int MAX = 3e3 + 41;
const int MOD = 998244353;

int n, x;
int p2[MAX];

int add(int a, int b) {
	a += b;
	if (a >= MOD) a -= MOD;
	return a;
}

int sub(int a, int b) {
	a -= b;
	if (a < 0) a += MOD;
	return a;
}

int mul(int a, int b) {
	return (ll) a * b % MOD;
}

int C[MAX][MAX];

void init() {
	fi(0, MAX - 1) {
		C[i][0] = 1;
		fj(1, i) {
			C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]);
		}		
	}
	p2[0] = 1;
	fi(1, MAX - 1) p2[i] = mul(p2[i - 1], 2);
}

int calcless() {
	int res = 0;	
	fi(0, n) {
		if (i >= x) break;
		fj(0, n) {
			if (i + j * 2 >= x) break;
			if (i + j > n) break;			
			int v1 = C[n][i + j];
			int v2 = C[i + j][i];
			int v = mul(v1, v2);
			res = add(res, v);
		}
	}
	return res;
}

int getways(int l) {	
	
	int res = 0;
	bool flag = 0;	
	for (int l0 = 2; l0 <= l; l0++) {
		if (l0 * 2 < x + 1) continue;
		int pref = l - l0 + 1;
		if (pref >= l0 - 1) {
			flag = 1;
			continue;
		}
		int q2 = pref + 1;
		int rem = l0 - q2;		
		int rem2 = x + 1 - q2 * 2 - rem;
		if (rem2 < 0) continue;
		if (rem2 > rem) continue;
		if (rem2 == rem) {
			flag = 1;
			continue;
		}
		int ways = C[rem][rem2];		
		res = add(res, ways);
	}
//	res = add(res, flag);
	int want = x + 1;
	if (want % 2 == 0) {
		if (want / 2 <= l) {
			res = add(res, 1);
		}
	}
	return res;
}

void solve () {
	if (x == 1) {
		printf("%d\n", p2[n]);
		return;
	}
	int ans = 0;
	ans = add(ans, calcless());
	err("less = %d\n", ans);
	fo(0, n) {
		int v1 = getways(o);
		int v2 = C[n][o];
		int v = mul(v1, v2);
		err("l = %d ways = %d\n", o, v);
		ans = add(ans, v);
	}	
	printf("%d\n", ans);
}

int main() {
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	START_TIME = (double)clock();
#endif

	init();
		
	scanf("%d %d", &n, &x);
	solve();

	exit();
	return 0;
}



