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>
using namespace std;
#define PII pair<int, int>
#define PLL pair<LL, LL>
#define VI vector<int>
#define VPII vector<PII>
#define LL long long
#define LD long double
#define f first
#define s second
#define MP make_pair
#define PB push_back
#define pb pop_back
#define ALL(c) (c).begin(), (c).end()
#define SIZ(c) (int)(c).size()
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define FOR(i, b, e) for(int i = (b); i <= (int)(e); ++i)
#define FORD(i, b, e) for(int i = (b); i >= (int)(e); --i)

#define Sim template<class n
Sim, class s> ostream & operator << (ostream &p, pair<n, s> x)
{return p << "<" << x.f << ", " << x.s << ">";}
Sim> auto operator << (ostream &p, n y) -> 
typename enable_if<!is_same<n, string>::value, decltype(y.begin(), p)>::type 
{int o = 0; p << "{"; for(auto c: y) {if(o++) p << ", "; p << c;} return p << "}";}
void dor() {cerr << endl;}
Sim, class...s> void dor(n p, s...y) {cerr << p << " "; dor(y...);}
Sim, class s> void mini(n &p, s y) {if(p>y) p = y;}
Sim, class s> void maxi(n &p, s y) {if(p<y) p = y;}
#ifdef DEB
#define debug(...) dor(__FUNCTION__, ":", __LINE__, ": ", __VA_ARGS__)
#else
#define debug(...)
#endif 

#define I(x) #x " = ", (x), " "
#define A(a, i) #a "[" #i " = ", i, "] = ", a[i], " "
const LL mod = 998244353;
const int M = 3003;

int n, x;
LL dp[M][2 * M], dp2[M][2 * M];
LL newt[M][M];

int main()
{
	//ios_base::sync_with_stdio(0);
	//cin.tie(NULL);
	FOR(i, 0, M - 1)
	{
		newt[i][0] = 1;
		FOR(j, 1, i)
		{
			if (i == 0)
			{
				break;
			}
			if (j > 0)
			{
				newt[i][j] += newt[i - 1][j - 1];
			}
			newt[i][j] += newt[i - 1][j];
			newt[i][j] %= mod;
		}
	}
	scanf("%d%d", &n, &x);
	LL res = 0;
	dp[0][0] = 1;
	dp2[0][0] = 1;
	FOR(i, 1, n)
	{
		FOR(j, 0, 2 * i)
		{
			dp[i][j] = dp[i - 1][j];
			if (j >= 1)
				dp[i][j] += dp[i - 1][j - 1];
			if (j >= 2)
				dp[i][j] += dp[i - 1][j - 2];
			dp[i][j] %= mod;
			if (j >= 1)
				dp2[i][j] += dp2[i - 1][j - 1];
			if (j >= 2)
				dp2[i][j] += dp2[i - 1][j - 2];
			dp2[i][j] %= mod;
		}
	}
	FOR(i, 0, 2 * n)
	{
		debug(i, i - 1, res);
		if (i < x)
		{
			res = (res + dp[n][i]) % mod;
			continue;
		}
		if (i == x)
		{
			continue;
		}
		int diff = i - x;
		if (diff % 2 == 0)
		{
			continue;
		}
		int boki = i - (x - 1);
		boki *= 2;
		if (boki >= i)
		{
			if (x % 2 == 1)
			{
				//res = (res + 1) % mod;
				res = (res + newt[n][i / 2]) % mod;
				debug("ololol", i, res, n, i / 2, newt[3][2]);
			}
			continue;
		}
		int l = n - boki / 2;
		FOR(j, 0, l)
		{
			res = (res + dp2[j][i - boki] * newt[n][boki / 2 + j]) % mod;
		}
	}
	printf("%lld\n", res);
	return 0;
}