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<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
typedef long double ld;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second

ll pas[6001][6001];
ll rpas[6002][6001];
ll dp[3001][6001];
void init(int sz) {
	rep(i, 2*sz+1) {
		rep(j, 2*sz+1) {
			if (i == 0 || j == 0)pas[i][j] = 1;
			else pas[i][j] = (pas[i - 1][j] + pas[i][j - 1]) % mod;
		}
	}
	rep(j, 2*sz+1) {
		rep(i, 2*sz+1) {
			rpas[i + 1][j] = (rpas[i][j] + pas[i][j]) % mod;
		}
	}
	dp[0][0] = 1;
	rep(i, sz) {
		rep(j, 2 * i + 1) {
			rep(k, 3) {
				dp[i + 1][j + k] += dp[i][j];
				if (dp[i + 1][j + k] >= mod)dp[i + 1][j + k] %= mod;
			}
		}
	}
}
void solve() {
	int n, x; cin >> n >> x;
	init(n);
	ll ans = 0;
	rep(i, x) {
		ans += dp[n][i];
		if (ans >= mod)ans -= mod;
	}
	//cout << ans << endl;
	/*if (x % 2) {
		for (int i = (x + 1) / 2; i <= n; i++) {
			ans += pas[n - i][i];
			if (ans >= mod)ans -= mod;
		}
	}*/
	//cout << ans << endl;
	rep1(k, n) {
		ll csum = 0;
		if (x % 2 && 2 * k > x) {
			csum++;
		}
		rep1(j, n - 1) {
			if (x-k < 0)continue;
			if (2 * k - x - 2 * j < 0)continue;
			csum += rpas[x-k][2 * k - x - 2 * j];
			if (csum >= mod)csum -= mod;
		}
		csum = csum * pas[n-k][k] % mod;
		ans += csum; if (ans >= mod)ans -= mod;
	}
	cout << ans << endl;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	//stop
		return 0;
}