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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll>pl;
const ll asize=3005,inf=1000000000000000000;
vector<ll> dp[asize],p2={1};
 
int main(){
	ios_base::sync_with_stdio(0),cin.tie(NULL);
	ll n,s,a;
	vector<ll>v;
	cin>>n>>s;
	for(int i=0;i<n;i++){
		cin>>a;
		v.push_back(a);
		ll k=p2.back()*2;
		k%=998244353;
		p2.push_back(k);
	}
	for(auto x:v){
		for(int i=s-x;i>=0;i--){
			for(auto j:dp[i]){
				dp[i+x].push_back(j+1);
			}
		}
		dp[x].push_back(1);
	}
	//for(auto x:p2)cout<<x<<" ";
	ll ans=0;
	for(auto x:dp[s]){
		ans+=p2[n-x];
		ans%=998244353;
	}
	cout<<ans;
/*	for(int i=0;i<100;i++){
		cout<<i<<": ";
		for(auto x:dp[i])cout<<x<<" ";
		cout<<"\n";
	}*/
	return 0;
}