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>
using namespace std;
using ll = long long;

const int inf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3fLL;

template <ll mod>
class modint
{
public:
    constexpr modint(ll value = 0) : _value(value >= 0 ? value % mod : (value % mod) + mod) {}
 
    constexpr modint operator+() const { return *this; }
    constexpr modint operator-() const { return modint(-_value); }
 
    constexpr modint &operator+=(const modint &rhs) {
        if ((_value += rhs._value) >= mod) _value -= mod;
        return *this;
    }
    constexpr modint &operator-=(const modint &rhs) {
        if ((_value += mod - rhs._value) >= mod) _value -= mod;
        return *this;
    }
    constexpr modint &operator*=(const modint &rhs) {
        (_value *= rhs._value) %= mod;
        return *this;
    }
    constexpr modint &operator/=(const modint &rhs) {
        *this *= rhs.inv();
        return *this;
    }
    constexpr modint operator+(const modint &rhs) const { return modint(*this) += rhs; }
    constexpr modint operator-(const modint &rhs) const { return modint(*this) -= rhs; }
    constexpr modint operator*(const modint &rhs) const { return modint(*this) *= rhs; }
    constexpr modint operator/(const modint &rhs) const { return modint(*this) /= rhs; }
 
    constexpr modint operator++() { *this += 1; return *this; }
    constexpr modint operator--() { *this -= 1; return *this; }
    constexpr modint operator++(int) { modint tmp = *this; ++*this; return tmp; }
    constexpr modint operator--(int) { modint tmp = *this; --*this; return tmp; }
 
    constexpr bool operator==(const modint &rhs) const { return _value == rhs._value; }
    constexpr bool operator!=(const modint &rhs) const { return _value != rhs._value; }
 
    constexpr modint inv() const {
        ll a = _value, b = mod, u = 1LL, v = 0LL;
        while (b > 0) {
            ll tmp = a / b;
            a -= tmp * b; swap(a, b);
            u -= tmp * v; swap(u, v);
        }
        return modint(u);
    }
 
    constexpr modint pow(ll n) const {
        modint e = *this, res = 1;
        while (n) {
            if (n & 1) res *= e;
            e *= e, n >>= 1;
        }
        return res;
    }
 
    friend ostream &operator<<(ostream &os, const modint &m) {
        os << m._value;
        return os;
    }
 
    friend istream &operator>>(istream &is, modint &m) {
        ll in;
        is >> in;
        m = modint(in);
        return is;
    }
 
private:
    ll _value;
};
using mint = modint<998244353>;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, s; cin >> n >> s;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];

    vector<mint> dp(s + 1);
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = s; j >= 0; j--) {
            dp[j] *= 2;
            if (j >= a[i]) dp[j] += dp[j - a[i]];
        }
    }
    cout << dp[s] << endl;
    return 0;
}