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 <algorithm>
#include <vector>
#include <string.h>
#include <cassert>
#include <numeric>
using namespace std;
using lint = long long;
const lint mod = 998244353;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl((lint)(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
lint power(lint base, lint exponent, lint module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ lint root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr first, itr last){ for(auto i = first; i != last; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<lint> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } lint C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };
template<class itr> bool next_sequence(itr first, itr last, int max_bound){ itr now = last; while(now != first){ now--; (*now)++; if((*now) == max_bound){ (*now) = 0; }else{ return true; } } return false; }

combination table(100000);

lint dp[3030][6060];

lint patterns(int cnt, int sum){
    if(sum <= 0){
        return !sum && !cnt;
    }else if(cnt == 0){
        return 0;
    }else if(dp[cnt][sum] != -1){
        return dp[cnt][sum];
    }else{
        return dp[cnt][sum] = (patterns(cnt - 1, sum - 1) + patterns(cnt - 1, sum - 2)) % mod;
    }
}

lint dp2[3030][6060];

lint patterns_colorful(int cnt, int sum){
    if(sum < 0){
        return 0;
    }else if(cnt == 0){
        return !sum;
    }else if(dp2[cnt][sum] != -1){
        return dp2[cnt][sum];
    }else{
        return dp2[cnt][sum] = (patterns_colorful(cnt - 1, sum) + patterns_colorful(cnt - 1, sum - 1) + patterns_colorful(cnt - 1, sum - 2)) % mod;
    }
}

int main(){
    int N, X;
    cin >> N >> X;
    lint ans = 0;
    memset(dp, -1, sizeof(dp));
    memset(dp2, -1, sizeof(dp2));
    for(int i = 0; i < X; i++){
        ans = (ans + patterns_colorful(N, i)) % mod;
    }
    for(int i = X + 1; i < 2 * X - 2; i++){
        if((i - X + 1) % 2 == 1){
            continue;
        }
        int remaining = N - (i - X + 1);
        if(remaining < 0){
            continue;
        }
        for(int j = 0; j <= remaining; j++){
            ans = (ans + patterns(j, 2 * X - 2 - i) * table.C(N, remaining - j) % mod) % mod;
        }
    }
    if(X % 2 == 1){
        for(int i = max(X + 1, 2 * X - 2); i <= 2 * N; i++){
            if(i % 2 == 1){
                continue;
            }
            ans = (ans + table.C(N, i / 2)) % mod;
        }
    }
    cout << ans << endl;
}