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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define all(v) v.begin(), v.end()
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define Rep(n) for (ll _ = 0; _ < (ll)(n); _++)
#define P pair<ll,ll>
#define fs first
#define sc second

template<class S>S sum(vector<S>&a){return accumulate(all(a),S());}
template<class S>S max(vector<S>&a){return *max_element(all(a));}
template<class S>S min(vector<S>&a){return *min_element(all(a));}
ll max(int a,ll b){return max((ll)a,b);} ll max(ll a,int b){return max(a,(ll)b);}
ll min(int a,ll b){return min((ll)a,b);} ll min(ll a,int b){return min(a,(ll)b);}
template <class S> void print(vector<S> &v){for(ll i=0; i<(ll)v.size(); i++){cerr << v[i] << ' ';}cerr << endl;}
template <class S> void print(vector<vector<S>> &v) {for (ll i = 0; i < (ll)v.size(); i++) {for (ll j = 0; j < (ll)v[i].size(); j++) {cerr << v[i][j] << ' ';}cerr << endl;}}

const int mod = 998244353;
struct mint {
  ll x;  // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

int main(){
	int n,s; cin >> n >> s;
	vector<int> a(n);
	rep(i,n) cin >> a[i];
	mint inv2 = mint(2).inv();
	vector<vector<mint>> dp(n, vector<mint>(s+1,mint(0)));
	dp[0][a[0]] = mint(2).pow(n-1);
	for(int i=1; i<n; i++){
		for(int j=1; j<=s; j++){
			dp[i][j] += dp[i-1][j];
			if(j - a[i] == 0){
				dp[i][j] += mint(2).pow(n-1);
			}else if(j - a[i] > 0){
				dp[i][j] += dp[i-1][j-a[i]] * inv2;
			}
		}
	}
	cout << dp[n-1][s] << endl;
	return 0;
}