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>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
 
using namespace std;

 
const int MAX_N = int(1e6) + 10;
long long int MOD=1e9+7;
const long long int INF=1e18;
 
typedef long long int ll;
typedef unsigned long long int llu;
 
 
 
inline ll mul(ll a,ll b){
 
    a=(a+MOD)%MOD;
    b=(b+MOD)%MOD;
    ll c=a*b;
    c=(c+MOD)%MOD;
    return c;
}
inline ll add(ll a,ll b){
    a=(a+MOD)%MOD;
    b=(b+MOD)%MOD;
    ll ans=(a+b);
    ans=(ans+MOD)%MOD;
    return ans;
}
ll power(ll a,ll b,ll c=0){
    ll ans=1;
    while(b){
        if(b&1){
            ans=mul(ans,a);
        }
        a=mul(a,a);
        b=b/2;
    }
    return ans;
}
 
  
 
 
#define all(x) (x).begin(),(x).end()




//s.reserve(32768); //updated !
//s.max_load_factor(0.25);



ll a[30005];
ll dp[30005];



int main(){
   
   int n,s;
   cin>>n>>s;


   MOD=998244353LL;

   dp[0]=power(2LL,n*1LL,MOD);
   ll inversess=power(2LL,MOD-2,MOD);

   for(int i=0;i<n;i++){
    cin>>a[i];
    for(int j=s;j>=a[i];j--){
        dp[j]=add(dp[j],mul(dp[j-a[i]],inversess));
    }
   }

   cout<<dp[s]<<endl;



   
   
    
}