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>
#define mp make_pair
#define fi first
#define se second
#define VAL 100000
#define ll  long long int
#define INF 10000000000000000LL
#define pll pair<long long , long long>
const int MAX=1e7;
const ll MOD=998244353;
const int TOT_PRIMES=19;
const int MAX_A=70;
using namespace std;
ll inv(ll a, ll b){
	ll res=1;
	while(b){
		if(b%2)
			res=(res*a)%MOD;
		b=b/2;
		a=(a*a)%MOD;
	}
	return res;
}
int main() 
{
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL);  
    cout.tie(0);
    int n,s;
    cin>>n>>s;
    ll dp[n+1][s+1];
    ll arr[n+1];
    for(int i=1;i<=n;i++)
    	cin>>arr[i];
    memset(dp,0,sizeof(dp));
   dp[0][0]=inv((ll)2, (ll)(n));
   for(int i=1;i<=n;i++){
   	for(int j=0;j<=s;j++){
   		dp[i][j]+=dp[i-1][j];
   		if(j+arr[i]<=s)
   			dp[i][j+arr[i]]+=(dp[i-1][j])*inv((ll)(2),MOD-2);
   		dp[i][j]%=MOD;
   	}
   }
   cout<<dp[n][s];
}