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;

typedef long long ll;
const int maxn=1e6+7;
const int inf=INT_MAX;
const ll inff=1e18;
const ll mod=998244353;
#define pii pair<int,int>
#define mkp make_pair
#define F first
#define S second
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
#define int ll

#ifdef HNO2
#define IOS
#else
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0);
#endif // HNO2

int ppow(int x,int y)
{
    if (y==0) return 1;
    int ret=ppow(x,y/2);
    ret=ret*ret%mod;
    if (y&1) ret=ret*x%mod;
    return ret;
}

int a[3007];
int dp[3007];

int32_t main()
{
    IOS
    int n,S;
    cin>>n>>S;
    for (int i=1;i<=n;i++) cin>>a[i];
    int inv2=ppow(2,mod-2);
    dp[0]=ppow(2,n);
    for (int i=1;i<=n;i++)
        for (int j=S;j>=0;j--)
            if (j>=a[i]) dp[j]=(dp[j]+dp[j-a[i]]*inv2%mod)%mod;
    cout<<dp[S]<<endl;
}