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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
typedef vector<double> VD;
typedef vector<VD> VVD;

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

const ll mod = 998244353;
const int N = 6010;
ll fact[N], invf[N];
ll dp[N][N];

ll add(ll x, ll y){
    return (x+y)%mod;
}

ll mul(ll x, ll y){
    return (x%mod)*(y%mod)%mod;
}

ll powll(ll x, ll y){
    ll res = 1LL;
    while(y){
        if (y & 1LL)
            res *= x;
        res %= mod;
        x = (x*x) % mod;
        y >>= 1LL;
    }
    return res;
}

ll divll(ll x, ll y){
    return (x * powll(y,mod-2)) % mod;
}

ll nPr(ll n, ll r){
    if (n < r || r < 0) return 0;
    return mul(fact[n], invf[n-r]);
}

ll nCr(ll n, ll r){
    if (n < r || r < 0) return 0;
    return mul(mul(fact[n], invf[r]), invf[n-r]);
}

int main() {
    ll n, x;
    cin >> n >> x;

    fact[0] = invf[0] = 1;
    FOR(i,1,N-1){
        fact[i] = (fact[i-1] * i) % mod;
        invf[i] = divll(invf[i-1], i);
    }

    dp[0][0] = 1;
    REP(i,n) REP(j,N-2){
        dp[i+1][j+1] = (dp[i+1][j+1] + dp[i][j]) % mod;
        dp[i+1][j+2] = (dp[i+1][j+2] + dp[i][j]) % mod;
    }

    ll ans = 1;
    FOR(m,1,n){
        // cout << "m " << m << endl;
        ll tmp = 0;
        REP(j,x) tmp = (tmp + dp[m][j]) % mod;
        FOR(t,1,(m-1)/2){
            ll d = x - 1 - 2 * t;
            if (d > 0 && d < 2 * (m - 2 * t)) tmp = (tmp + dp[m - 2 * t][d]) % mod;
            // if (2 * t == x) break;
            // ll nn = m - 2 * t;
            // for (ll xx = x - 3; xx > 0 && xx + 4 * t >= x; xx -= 2){
            //     ll d = 0;
            //     if (nn == 1){
            //         if (xx == 1) d = 1;
            //     }else{
            //         d = dp[nn][xx];
            //         if (xx >= 4) d = (d - dp[nn-2][xx-4]) % mod;
            //     }
            //     tmp += d;
            //     cout << nn << " " << xx << " " << d << endl;
            // }
            // cout << d << " " << m - 2 * t << " " << dp[m - 2 * t][d] << endl;
        }
        if (x % 2 == 1 && 2 * m > x) tmp++;
        // cout << tmp << endl;
        ans = (ans + tmp * nCr(n, m)) % mod;
    }
    cout << ans << endl;

    return 0;
}
