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;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define INF 1000000000
#define mod 998244353
using ll=long long;
const ll LINF=1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
// ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return true;}return false;}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    
    int n,s;cin>>n>>s;
    ll a[n];
    rep(i,n) cin>>a[i];

    ll dp[3][n+1][s+1];
    rep(i,3)rep(j,n+1)rep(k,s+1) dp[i][j][k]=0;

    // 0 not in not use, 1 in not use, 2 in use
    dp[0][0][0]=1;

    rep(i,n){
        for(int j=0;j<=s;j++){
            // use
            if(j-a[i]>=0){
                dp[2][i+1][j]=(dp[2][i+1][j]+dp[0][i][j-a[i]]+dp[1][i][j-a[i]]+dp[2][i][j-a[i]])%mod;
            }
            // not in not use
            dp[0][i+1][j]=(dp[0][i+1][j]+dp[0][i][j]+dp[1][i][j]+dp[2][i][j])%mod;
            dp[1][i+1][j]=(dp[1][i+1][j]+dp[0][i][j]+dp[1][i][j]+dp[2][i][j])%mod;
            
        }
    }

    // cout<<"0"<<endl;
    // rep(j,s+1){
    //     rep(i,n+1){
    //         cout<<dp[0][i][j]<<" ";
    //     }
    //     cout<<endl;
    // }

    // cout<<"1 -- "<<endl;
    // rep(j,s+1){
    //     rep(i,n+1){
    //         cout<<dp[1][i][j]<<" ";
    //     }
    //     cout<<endl;
    // }
    // cout<<"2 --- "<<endl;
    // rep(j,s+1){
    //     rep(i,n+1){
    //         cout<<dp[2][i][j]<<" ";
    //     }
    //     cout<<endl;
    // }

    ll ans=0;
    rep(i,3){
        ans+=dp[i][n][s];
        ans%=mod;
    }
    cout<<ans<<endl;
    return 0;
}