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 rep(i,n) for (int i = 0; i < (n); ++i)
#define chmax(a, b) a = max(a, b);
#define chmin(a, b) a = min(a, b);
using namespace std;
using ll = long long;
using P = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;

//#define MOD 1000000007
#define MOD 998244353
#define mod 998244353
struct mint {
	int i;
	mint() : i(0) {}
	mint(int x) {
		i = int(x % MOD);
		if (i < 0) i += MOD;
	}
	template<class T> mint(T x) {
		i = int(x % MOD);
		if (i < 0) i += MOD;
	}
	mint operator+(const mint x) const {return i + x.i;}
	mint operator-(const mint x) const {return i - x.i;}
	mint operator*(const mint x) const {return (long long)i * x.i;}
	mint operator/(const mint x) const {return (long long)i * x.pow(MOD - 2).i;}
	mint inv() {return pow(MOD - 2);}
	template<class T> mint pow(T p) const {
		long long r = 1;
		long long t = i;
		for(; p > 0; p >>= 1) {
			if (p & 1) r = r * t % MOD;
			t = t * t % MOD;
		}
		return r;
	}
	template<class T1, class T2> static mint pow(T1 a, T2 b) {
		long long r = 1;
		long long t = (long long)(a % MOD);
		for(; b > 0; b >>= 1) {
			if (b & 1) r = r * t % MOD;
			t = t * t % MOD;
		}
		return r;
	}
	mint& operator+=(const mint x) {
		i = (i + x.i) % MOD;
		return *this;
	}
	mint& operator-=(const mint x) {
		i = i - x.i;
		if (i < 0) i += MOD;
		return *this;
	}
	mint& operator*=(const mint x) {
		i = (int)((long long)i * x.i % MOD);
		return *this;
	}
	mint& operator/=(const mint x) {
		i = (long long)i * x.pow(MOD - 2).i % MOD;
		return *this;
	}
};

std::ostream& operator<<(std::ostream& os, const mint& m) {
	return os << m.i;
}

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} comb(7000);

int main() {
    int n, x;
    cin >> n >> x;
    mint ans(0);
    {
        vector<mint> dp(2*n+1);
        dp[0] = 1;
        rep(_, n) {
            for(int i = 2*n-2; i >= 0; i--) {
                dp[i+2] += dp[i];
                dp[i+1] += dp[i];
            }
        }
        for(int i = 0; i < x - 1; i++) ans += dp[i];
    }
    {
        vector<mint> dp(2*n+1);
        dp[0] = 1;
        for(int k = 1; k <= n; k++) {
            for(int i = 2*n-2; i >= 0; i--) {
                dp[i+2] += dp[i];
                dp[i+1] += dp[i];
                dp[i] = 0;
            }
            for(int t = x - 1; t >= 0; t -= 2) {
                int c = (x - 1 - t) / 2;
                int call = c * 2 + k;
                ans += dp[t] * comb(n, call);
            }
        }
    }
    if (x & 1) {
        for(int c = x - 1; c <= n; c++) {
            ans += comb(n, c);
        }
    }
    cout << ans << endl;
}
