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 i64 long long int
i64 m[3001][3001];
i64 mp=998244353;
int a[3001];
i64 f(int i,int s)
{
    if((i==0)&&(s==0)) return 1;
    if(i==0) return 0;
    if(m[i][s]!=-1) return m[i][s];
    if(s-a[i]>=0)
        m[i][s]=((2*f(i-1,s))%mp+f(i-1,s-a[i]))%mp;
    else m[i][s]=(2*f(i-1,s))%mp;
    return m[i][s];
}

int main()
{
   int n,s,i,j;
   i64 b;
   scanf("%d %d",&n,&s);
   for(i=1;i<=n;i++) scanf("%d",&a[i]);
   for(i=0;i<=n;i++)
   {
       for(j=0;j<=s;j++)
        m[i][j]=-1;
   }
   b=f(n,s);
   printf("%lld",b);
    return 0;
}