Solve in Python:
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-----
3 4
2 2 4

-----Sample Output-----
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\})
#!python3

iim = lambda: list(map(int, input().rstrip().split()))

def resolve():
    N, S = iim()
    A = list(iim())
    mod = 998244353
    S1 = S + 1

    dp = [0] * (S+1)

    dp[0] = pow(2, N, mod)
    inv = pow(2, mod-2, mod)

    for ai in A:
        for i in range(S, ai-1, -1):
            dp[i] = (dp[i] + dp[i-ai]*inv) % mod

    print((dp[-1]))


def __starting_point():
    resolve()

__starting_point()