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>

#define int long long

#define override_rep(i, l, r, mes, ...) mes
#define rep1(i, n) for (int i = 0; i < n; i++)
#define rep2(i, l, r) for (int i = l; i < r; i++)
#define rep(...) override_rep(__VA_ARGS__, rep2, rep1)(__VA_ARGS__)

using namespace std;
using P = pair<int, int>;

template<typename T, typename U>
istream& operator>>(istream& in, pair<T, U>& p) {
    in >> p.first >> p.second;
    return in;
}

template<int64_t MOD>
class ModInt {
	typedef int64_t T;
	T x;

  public:
	ModInt(T value) {
		if (value < 0) value = (MOD - (-value) % MOD);
		x = value % MOD;
	}

	ModInt operator-() const {
		return ModInt(-x);
	}

	ModInt& operator+=(const ModInt& p) {
		if ((x += p.get()) >= MOD) x -= MOD;
		return *this;
	}

	ModInt& operator-=(const ModInt& p) {
		if ((x += MOD - p.get()) >= MOD) x -= MOD;
		return *this;
	}

	ModInt& operator*=(const ModInt& p) {
		x = x * p.get() % MOD;
		return *this;
	}

	ModInt& operator/=(const ModInt& p) {
		*this *= p.inverse();
		return *this;
	}

	ModInt& operator+=(const T& p) {
		return *this += ModInt(p);
	}

	ModInt& operator-=(const T& p) {
		return *this -= ModInt(p);
	}

	ModInt& operator*=(const T& p) {
		return *this *= ModInt(p);
	}

	ModInt& operator/=(const T& p) {
		return *this /= ModInt(p);
	}

	ModInt operator+(const ModInt& p) const {
		return ModInt(*this) += p;
	}

	ModInt operator-(const ModInt& p) const {
		return ModInt(*this) -= p;
	}

	ModInt operator*(const ModInt& p) const {
		return ModInt(*this) *= p;
	}

	ModInt operator/(const ModInt& p) const {
		return ModInt(*this) /= p;
	}

	ModInt operator+(const T& p) const {
		return ModInt(*this) += p;
	}

	ModInt operator-(const T& p) const {
		return ModInt(*this) -= p;
	}

	ModInt operator*(const T& p) const {
		return ModInt(*this) *= p;
	}

	ModInt operator/(const T& p) const {
		return ModInt(*this) /= p;
	}

	bool operator==(const ModInt& p) const {
		return x == p.get();
	}

	bool operator!=(const ModInt& p) const {
		return x != p.get();
	}

	bool operator==(const T& p) const {
		return x == ModInt(p);
	}

	bool operator!=(const T& p) {
		return x != ModInt(p);
	}

	ModInt inverse() const {
		int a = x, b = MOD, u = 1, v = 0;
		while (b) {
			int t = a / b;
			a -= t * b;
			u -= t * v;
			swap(a, b);
			swap(u, v);
		}
		return ModInt(u);
	}

	T get() const{
		return x;
	}

	friend ostream& operator<<(ostream& out,const ModInt& value){
	    out<<value.x;
	    return out;
	}
};

// https://twitter.com/vane11ope ←フォローしましょう
// (ノ)>◡<(ヾ)ﾓﾁﾓﾁ
signed main(signed argc, char* argv[]) {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    const int MOD = 998244353;

    int n,s;
    cin>>n>>s;
    using moi=ModInt<MOD>;
    vector<int> a(n);
    rep(i,n)cin>>a[i];
    vector<vector<moi>> dp(n+1,vector<ModInt<MOD>>(s+1,ModInt<MOD>(0)));
    dp[0][0]=moi(1);
    rep(i,n){
        rep(j,s+1){
            dp[i+1][j]=dp[i][j]*2;
            if(j-a[i]>=0)dp[i+1][j]+=dp[i][j-a[i]];
        }
    }
    cout<<dp[n][s]<<endl;
}