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 rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,b,a) for(int i=b;i>=a;i--)
#define fori(a) for(auto i : a )
#define all(a) begin(a), end(a)
#define set(a,b) memset(a,b,sizeof(a))
#define sz(a) a.size()
#define pi 3.14159
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PF push_front //deque
#define mp make_pair
#define pq priority_queue
const ll mod = 998244353;
#define f first
#define s second
#define pii pair< int, int >
#define vi vector<int>
#define vpii vector<pii>
#define debug(v) for(auto i:v) cout<<i<<" ";
#define tc int t; cin >> t; while(t--)

using namespace std;
string repeat(string s, int n) {
    string s1 = "";
    for (int i=0; i<n;i++)
        s1+=s;
    return s1;
}
string getString(char x) {
    string s(1, x);
    return s;
}

void optimizeIO(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
int gcd(int a, int b){
    if (a == 0)  return b;
    return gcd(b % a, a);
}

void solve(){

  int  n,m;
  cin>>n>>m;
  int a[n+1];
  rep(i,1,n+1) cin>>a[i];
  vector<int> dp(3010,0);
  dp[0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = m; j >= 0; j--){
      // 2. put to T,but do not put to U
      // 3. do not put to both U and T
			dp[j] = dp[j]*2%mod;
      // 1. put to both U and T
			if (j-a[i] >= 0) dp[j] = (dp[j]+dp[j-a[i]])%mod;
		}
	}
	printf("%lld\n", dp[m]);

}
int main(){
    optimizeIO();
    {
      solve();
    }
}