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 int lld;
typedef pair<int,int> pi;
typedef pair<lld,lld> pl;
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define sz(x) (int)x.size()
#define mk(a,b) make_pair(a,b)

const lld mod =  998244353;
int arr[3333];
lld dp[3333];
int dp2[2][3333][3333];
int main(void){
	dp[0] = 1;
	for(int e=1;e<3333;e++) dp[e] = (dp[e-1]*(lld)2)%mod;
	int n,s;
	scanf("%d%d",&n,&s);
	for(int p=0;p<n;p++) scanf("%d",&arr[p]);
	dp2[0][arr[0]][1] = 1;
	for(int e=1;e<n;e++){
		for(int p=0;p<=s;p++){
			for(int q=1;q<=n;q++){
				dp2[(e%2)][p][q] = dp2[(e+1)%2][p][q];
			}
		}
		dp2[(e%2)][arr[e]][1]++;
		for(int p=0;p<=s;p++){
			if(p+arr[e]>s) break;
			for(int q=1;q<=n;q++){
				dp2[(e%2)][p+arr[e]][q+1] += dp2[(e+1)%2][p][q];
			}
		}
	}
	lld tot = 0;
	for(int e=1;e<=n;e++){
		 tot = (tot+dp2[(n-1)%2][s][e]*dp[n-e])%mod;
	}
	printf("%lld\n",tot);
	return 0;
}