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>
#include <unistd.h>

using namespace std;

#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}


typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
//const ll MOD = 1000000007;
const ll MOD = 998244353;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { -1,0, 1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
//#define mp make_pair

const int MAXN = 6010;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = fact[1] = rfact[0] = rfact[1] = 1;
  for (ll i = 2; i <= n; i++) {
    fact[i] = i * fact[i - 1] % M;
    rfact[i] = mod_inverse(fact[i], M);
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
//nが大きくfactが計算できないときのほかの計算方法について書いてある
ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  ll ret = fact[n];
  ret = (ret*rfact[r]) % M;
  ret = (ret*rfact[n - r]) % M;
  return ret;
}



int main() {
  int n, x;
  cin >> n >> x;
  set_fact(max(x, 2*n) + 5);
  
  vl ans(n + 1);
  for (int k = 0; k <= n; k++) {
    for (int i = 0; i <= min(x - 1 - k, k); i++) {
      (ans[k] += nCr(k, i)) %= MOD;
    }
  }


  // 長さi, 合計j, 全部2以外
  vll dp(n + 1, vl(2*n + 1));
  dp[0][0] = 1;
  rep (i, n) {
    rep (j, 2*n + 1) {
      for (int k = 1; k <= 2; k++) {
        if (j + k <= 2*n) {
          (dp[i + 1][j + k] += dp[i][j]) %= MOD;
        }
      }
    }
  }
  rep (i, n + 1) {
    dp[i][2*i]--;
  }

  for (int s = x + 1; s < 2*x; s++) {
    if ((s - x + 1) % 2 == 1) continue;
    int num_i = (s - x + 1) / 2;
    int len = 2*num_i, sum = 2*2*num_i;
    for (int k = 0; k <= n; k++) {
      if (k < len || s < sum) continue;
      (ans[k] += dp[k - len][s - sum]) %= MOD;
    }
  }

  if (x % 2 == 1) {
    for (int s = x + 1; s <= 2 * n; s += 2) {
      (ans[s / 2] += 1) %= MOD;
    }
  }
  
  ll res = 0;
  rep (i, n + 1) {
    int nn = i + 1, rr = n - i;
    ans[i] *= nCr(nn + rr - 1, rr);
    ans[i] %= MOD;
    (res += ans[i]) %= MOD;
  }
  cout << res << endl;
}
