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	INF	0x3f3f3f3f
#define	LL	long long
#define	MAXN	3010
#define	mod	998244353
using namespace std;
int n, x, ans;
int fac[MAXN], inv[MAXN];

template <typename T> void chkmin(T &x, T y){x = min(x, y);}
template <typename T> void chkmax(T &x, T y){x = max(x, y);}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}

int mul(const int &x, const int &y){return 1ll * x * y % mod;}
int inc(const int &x, const int &y){return (x + y >= mod) ? (x + y - mod) : (x + y);}
int dec(const int &x, const int &y){return (x - y < 0) ? (x - y + mod) : (x - y);}
void update(int &x, const int &y){x = (x + y >= mod) ? (x + y - mod) : (x + y);}

int qpow(int a, int b){
	int ret = 1;
	while (b) {
		if (b & 1) ret = mul(ret, a);
		a = mul(a, a);
		b >>= 1;
	}
	return ret;
}

void init(int n){
	fac[0] = inv[0] = 1;
	for (int i = 1; i <= n; ++i)
		fac[i] = mul(fac[i - 1], i);
	inv[n] = qpow(fac[n], mod - 2);
	for (int i = n - 1; i >= 1; --i)
		inv[i] = mul(inv[i + 1], i + 1);
}

int C(int n, int m){
	return mul(fac[n], mul(inv[m], inv[n - m]));
}

int main(){
	read(n), read(x);
	init(n);
	ans = 1;
	for (int s = 1; s < x; ++s){
		for (int len = s / 2 + s % 2; len <= n && len <= s; ++len)
			update(ans, mul(C(len, s - len), C(n, len)));
	}
	for (int s = x + 1; s <= 2 * n; s += 2){
		int i = (s - (x - 1)) / 2;
		if (2 * i >= x - 1) {
			if (x & 1) {
				int len = s / 2;
				update(ans, C(n, len));
			}
		} else {
			int ss = x - 1 - 2 * i;
			for (int len = ss / 2 + ss % 2; len + 2 * i <= n && len <= ss; ++len)
				update(ans, mul(C(len, ss - len), C(n, len + 2 * i)));
		}
	}
	printf("%d\n", ans);
	return 0;
}