Problem:
Score : 600 points

Problem Statement

Given are a sequence of N positive integers A_1, A_2, \ldots, A_N and another positive integer S.

For a non-empty subset T of the set \{1, 2, \ldots , N \}, let us define f(T) as follows:

f(T) is the number of different non-empty subsets \{x_1, x_2, \ldots , x_k \} of T such that A_{x_1}+A_{x_2}+\cdots +A_{x_k} = S.

Find the sum of f(T) over all 2^N-1 subsets T of \{1, 2, \ldots , N \}. Since the sum can be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N \leq 3000

1 \leq S \leq 3000

1 \leq A_i \leq 3000

Input

Input is given from Standard Input in the following format:

N S
A_1 A_2 ... A_N

Output

Print the sum of f(T) modulo 998244353.

Sample Input 1

3 4
2 2 4

Sample Output 1

6

For each T, the value of f(T) is shown below. The sum of these values is 6.

f(\{1\}) = 0

f(\{2\}) = 0

f(\{3\}) = 1 (One subset \{3\} satisfies the condition.)

f(\{1, 2\}) = 1 (\{1, 2\})

f(\{2, 3\}) = 1 (\{3\})

f(\{1, 3\}) = 1 (\{3\})

f(\{1, 2, 3\}) = 2 (\{1, 2\}, \{3\})

Sample Input 2

5 8
9 9 9 9 9

Sample Output 2

0

Sample Input 3

10 10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

3296

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <omp.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;


#define int ll
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());


typedef pair<int, int> pii;
typedef pair<string, string> pst;
typedef pair<pii, pii> piii;
typedef vector<piii> vpii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<pii> vpi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
typedef pair<pld, pld> pldd;
typedef vector<pld> vpd;
typedef complex<double> base;


#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define my_abs(a) (((a) >= 0) ? (a) : -(a))
#define mp(a, b) make_pair(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define MODIK(a) (a >= M ? a - M : a)
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cout << (#x) << " = " << x << endl;
#ifdef ART_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif


const double PI = 2 * acos(0.0);

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

istream &operator >> (istream &in, pii &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pii &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


istream &operator >> (istream &in, pld &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pld &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) <<
               ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void print(int a, int b, int n, vi h)
{
    ofstream cout("output.txt");
    cout << SZ(h) << ' ' << a << ' ' << b << '\n';
    for (int i = 0; i < SZ(h); ++i) {
        cout << h[i]  << ' ';
    }
    cout.close();
    exit(0);
}


inline bool break_time(ld time)
{
    return clock() > CLOCKS_PER_SEC * time;
}


ld get_rand_ld(ld l, ld r)
{
    uniform_real_distribution<double> gen(l, r);
    return gen(rng);
}


bool may(int E, ld T)
{
    if (E < 0)
        return 1;

    double eps = get_rand_ld(0, 1);
    return eps < exp(-E / T);
}


void smain();

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

#ifdef ART_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(0) << fixed;
    smain();
#ifdef ART_HOME
    cout << "\nTOTAL EXECUTIOn TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
}


const int N = 3000 + 10, K = 1e2 + 10, B = 1000, oo = 1e18, M = 998244353, LOG = 20;
const int M1 = 1000000093, x1 = 27162;
const int M2 = 1000000087, x2 = 241;
const ld pi = atan2(1, 0) * 2, EPS = 1e-9;


int dp[N][N];


int MOD(int v)
{
    v %= M;
    if (v < 0)
        v += M;

    return v;
}


void smain() {
    int n, s;
    cin >> n >> s;

    dp[0][0] = 1;
    for (int i = 0; i < n; ++i)
    {
        int v;
        cin >> v;
        for (int j = 0; j + v <= s; ++j)
            dp[i + 1][j + v] = MOD(dp[i + 1][j + v] + dp[i][j]);

        for (int j = 0; j <= s; ++j)
            dp[i + 1][j] = MOD(dp[i + 1][j] + dp[i][j] * 2);
    }

    cout << dp[n][s];
}