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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define endl "\n"

const i64 MOD = 998244353;

int main()
{
  i64 N, S;
  cin >> N >> S;
  vector<i64> A(N);
  for (i64 i = 0; i < N; i++)
    cin >> A[i];
  vector<vector<i64>> dp(N + 1, vector<i64>(S + 1));
  dp[0][0] = 1;
  for (i64 i = 0; i < N; i++)
  {
    for (i64 j = 0; j <= S; j++)
      (dp[i + 1][j] = dp[i][j] * 2) %= MOD;
    for (i64 j = 0; j <= S - A[i]; j++)
      (dp[i + 1][j + A[i]] += dp[i][j]) %= MOD;
  }
  cout << dp[N][S] << endl;
  return 0;
}