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
const long long MOD = 998244353;
const int N = 3004;

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <unordered_set>
#include <bitset>
#include <deque>

using namespace std;

#define int long long
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<ll> vi;

#define forn(i, n) for (int (i) = 0; (i) != (n); (i)++)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define fi first
#define se second
#define re return
#define uniq(x) sort(all(x)); (x).resize(unique(all(x)) - (x).begin())

#ifdef LOCAL
#define dbg(x) cerr << __LINE__ << " " << #x << " " << x << endl
#define ln cerr << __LINE__ << endl
#else
#define dbg(x) void(0)
#define ln void(0)
#endif // LOCAL

int cx[4] = {-1, 0, 1, 0};
int cy[4] = {0, -1, 0, 1};

ll inq(ll x, ll y)
{
    if (!y) re 1 % MOD;
    ll l = inq(x, y / 2);
    if (y % 2) re l * l % MOD * x % MOD;
    re l * l % MOD;
}

ll rev(ll x)
{
    return inq(x, MOD - 2);
}

bool __precomputed_combinatorics = 0;
vector<ll> __fact, __ufact, __rev;

void __precompute_combinatorics()
{
    __precomputed_combinatorics = 1;
    __fact.resize(N);
    __ufact.resize(N);
    __rev.resize(N);
    __rev[1] = 1;
    for (int i = 2; i < N; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;
    __fact[0] = 1, __ufact[0] = 1;
    for (int i = 1; i < N; i++) __fact[i] = __fact[i - 1] * i % MOD, __ufact[i] = __ufact[i - 1] * __rev[i] % MOD;
}

ll fact(int x)
{
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[x];
}

ll cnk(int n, int k)
{
    if (k < 0 || k > n) return 0;
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;
}

int n, x;

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> x;
    vector<ll> calc(n + 1);
    calc[0] = 1;
    for (int len = 1; len <= n; len++)
    {
        for (int summ = len; summ <= 2 * len; summ++)
        {
            if (summ < x)
            {
                calc[len] = (calc[len] + cnk(len, summ - len)) % MOD;
            }
            else
            {
                if (summ <= x || (summ - x + 1) % 2 == 1)
                {
                    continue;
                }
                int L = (summ - x + 1) / 2;
                int R = len - L;
                int summa = summ - 4 * L;
                if (summa + 2 * L != x - 1) cout << "My bad\n", exit(0);
                int dlina = len - 2 * L;
                //cout << len << " " << summ << " " << L << " " << dlina << endl;
                if (summa != 2 * dlina) calc[len] = (calc[len] + cnk(dlina, summa - dlina)) % MOD;
            }
        }
        if (2 * len >= x && x % 2 == 1) calc[len] = (calc[len] + 1) % MOD;
        //cout << len << " " << calc[len] << endl << endl;
    }
    ll ans = 0;
    for (int i = 0; i <= n; i++)
    {
        ans = (ans + calc[i] * cnk(n, i)) % MOD;
    }
    cout << ans;
}

/* Note:
Check constants at the beginning of the code.
    N is set to 4e5 but be careful in problems with large constant factor.
    Setting N in every problem is more effective.
Check corner cases.
    N = 1
No def int long long for now.
Add something here.
*/
