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;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); ++i)
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define fi first
#define se second
#define INF 1000000009
#define LLINF 1000000000000000009LL
using ll = long long;

const int mod = 998244353;
struct mint {
  long long x;
  mint(long long 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(long long 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;
}

mint dp[3009][3009];

int main(){
  int n,s;
  cin>>n>>s;
  vector<int>a(n);
  rep(i,n)cin>>a[i];

  dp[0][0]=1;
  rep(i,n){
    rep(j,s+1){
      dp[i+1][j]+=dp[i][j]*2;
      if(j+a[i]<=s){
        dp[i+1][j+a[i]]+=dp[i][j];
      }
    }
  }

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

  return(0);
}