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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

ll MOD = 1e9+7;

class mint{
    ll n;
public:
    mint(ll m = 0){
        n = ((m % MOD) + MOD) % MOD;
    }
    mint operator-() const {
        return mint(-n);
    }
    mint& operator+=(const mint& m){
        n += m.n;
        if(n >= MOD) n -= MOD;
        return *this;
    }
    mint& operator-=(const mint& m){
        n += MOD - m.n;
        if(n >= MOD) n -= MOD;
        return *this;
    }
    mint& operator*=(const  mint& m){
        n *= m.n;
        n %= MOD;
        return *this;
    }
    mint operator+(const mint& m) const{
        mint res(*this);
        res += m;
        return res;
    }
    mint operator-(const mint& m) const{
        mint res(*this);
        res -= m;
        return res;
    }
    mint operator*(const mint& m) const{
        mint res(*this);
        res *= m;
        return res;
    }
    mint pow(ll t) const{
        if(t == 0) return mint(1);
        mint m = pow(t>>1);
        m *= m;
        if(t & 1 == 1) m *= *this;
        return m;
    }
    // for prime modulo
    mint inv() const{
        return pow(MOD-2);
    }
    mint& operator/=(const mint& m){
        (*this) *= m.inv();
        n %= MOD;
        return *this;
    }
    mint operator/(const mint& m) const{
        mint res(*this);
        res /= m;
        return res;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.n;
        return os;
    }
};

ll N,S;
ll A[3000];
mint dp[3050][3050];

int main(){
    cin >> N >> S;
    for(int i=0;i<N;i++) cin >> A[i];
    dp[0][0] = mint(2).pow(N);
    for(int i=0;i<N;i++){
        for(int j=0;j<=S;j++){
            dp[i+1][j] += dp[i][j];
            if(j+A[i]<=S){
                dp[i+1][j+A[i]] += dp[i][j] / mint(2);
            }
        }
    }
    cout << dp[N][S] << endl;
    return 0;
}