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 <stdio.h>
#define N 3000
#define S 3000
#define MOD 998244353
long long dp[N+1][S+1];
int a[N];

int main(){
  int n,s;
  int i,j;
  scanf("%d %d", &n, &s);
  for(i=0;i<n;i++)scanf("%d", &a[i]);

  dp[0][0]=1;

  for(i=0;i<n;i++){
    for(j=0;j<=S;j++){
      dp[i+1][j] += 2*dp[i][j];//tにi個選ぶ時それぞれのiについてuに選ぶ選ばないの2通りなのでこれで良い
      dp[i+1][j]%=MOD;
      if(j+a[i] <= S){
        dp[i+1][j+a[i]] += dp[i][j];
        dp[i+1][j+a[i]]%=MOD;
      }
    }
  }


    printf("%lld\n", dp[n][s]);

  return 0;
}