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>
#define llint long long
#define mod 998244353

using namespace std;

llint n, x;
llint dp[3005][6005];
llint cnt[3005];

const int FACT_MAX = 3005;
llint fact[FACT_MAX], fact_inv[FACT_MAX];

llint modpow(llint a, llint n)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2) % mod;
	}
}

void make_fact()
{
	llint val = 1;
	fact[0] = 1;
	for(int i = 1; i < FACT_MAX; i++){
		val *= i;
		val %= mod;
		fact[i] = val;
	}
	fact_inv[FACT_MAX-1] = modpow(fact[FACT_MAX-1], mod-2);
	for(int i = FACT_MAX-2; i >= 0; i--){
		fact_inv[i] = fact_inv[i+1] * (i+1) % mod;
	}
}

llint comb(llint n, llint k)
{
	llint ret = 1;
	ret *= fact[n];
	ret *= fact_inv[k], ret %= mod;
	ret *= fact_inv[n-k], ret %= mod;
	return ret;
}

int main(void)
{
	cin >> n >> x;
	make_fact();
	
	dp[0][0] = 1;
	for(int i = 0; i < n; i++){
		for(int j = 0; j <= 2*n; j++){
			for(int k = 1; k <= 2; k++){
				if(j+k <= 2*n){
					dp[i+1][j+k] += dp[i][j];
					dp[i+1][j+k] %= mod;
				}
			}
		}
	}
	
	for(int i = 0; i <= n; i++){
		for(int j = 0; j < x; j++){
			cnt[i] += dp[i][j], cnt[i] %= mod;
		}
	}
	for(int i = 1; i <= n; i++){
		for(int j = 0; j < i; j++){
			if(2*j-i <= 0) continue;
			if((x-1)-2*(i-j) <= 0) continue;
			if((x-1)-2*(i-j) >= 2*(2*j-i)) continue;
			cnt[i] += dp[2*j-i][(x-1)-2*(i-j)];
			cnt[i] %= mod;
		}
		if(x % 2 && 2*i > x) cnt[i]++, cnt[i] %= mod;
	}
	
	llint ans = 0;
	for(int i = 0; i <= n; i++){
		ans += cnt[i] * comb(n, i) % mod;
		ans %= mod;
	}
	cout << ans << endl;
	
	return 0;
}