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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

public class Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;
    long inv2 =1;
    public static void main(String[] args) throws IOException {
        Main main = new Main();
        main.solve();
//        out.println(main.solve());
        out.flush();
    }
    void solve() throws IOException {
        int n = sc.nextInt(), S = sc.nextInt();
        int[] a = new int[n];
        for(int i=0; i<n; i++) a[i] = sc.nextInt();
        long[][] dp = new long[n][S+1];
        inv2 = power(2, mod-2);
        if(a[0]<=S) dp[0][a[0]] = power(2, n-1);
        for(int i=1; i<n; i++){
            int cur = a[i];
            for(int j=1; j<=S; j++) dp[i][j] = dp[i-1][j];
            if(cur<=S){
                dp[i][cur] = (dp[i][cur]+power(2,n-1))%mod;
            }
            for(int j=cur+1; j<=S; j++){
                dp[i][j] += dp[i-1][j-cur]*inv2%mod;
                dp[i][j] %= mod;
            }
        }
        long ans = dp[n-1][S];
        out.println(ans);
    }
    long power(long base, long p){
        long ans = 1;
        while(p>0){
            if(p%2==1) ans = ans*base%mod;
            base = base*base%mod;
            p /= 2;
        }
        return ans;
    }
}