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 Scala (2.13.1) (judge verdict: Time Limit Exceeded)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, s = sc.nextInt()
  val a = IndexedSeq.fill(n)(sc.nextInt())

  val mod = 998244353
  def plus(a: Long, b: Long): Long = (a + b) % mod
  def times(a: Long, b: Long): Long = (a * b) % mod

  def pow(a: Long, n: Int, res: Long = 1): Long = {
    if (n == 0) {
      res
    } else if ((n&1) == 0) {
      pow(times(a,a), n >> 1, res)
    } else {
      pow(times(a,a), n >> 1, times(res,a))
    }
  }

  def inv(a: Long): Long = pow(a, mod-2)
  def div(a: Long, b: Long): Long = times(a, inv(b))

  val dp = Array.fill((n+1) * (s+1))(-1L)
  dp(0) = pow(2, n)
  for (j <- 1 to s) dp(j) = 0

  def solve(i: Int, j: Int): Long = {
    if (dp(i * (s+1) + j) != -1L) {
      dp(i * (s+1) + j)
    } else {
      val ans = if (j >= a(i-1)) {
        plus(solve(i-1, j), div(solve(i-1, j - a(i-1)), 2))
      } else {
        solve(i-1, j)
      }
      dp(i * (s+1) + j) = ans
      ans
    }
  }

  println(solve(n,s))
}