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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>

using namespace std;

#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)

using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=998244353 ;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll n, s;
    cin >> n >> s;
    vector<ll> a(n);//aは0-origin
    for (ll i=0; i<n; i++) {
        cin >> a[i];
    }
    vector<vector<ll>> dp(n+1, vector<ll>(s+1, 0)); // dp[i][j] : Aiまで考えたときTとUの積集合の要素を添え字としたaの和がjであるような場合の数 dpのiは1-origin
    dp[0][0] = 1;

    for (int i=0; i<n; i++) {
        for (int j=0; j<s+1; j++) {
            dp[i+1][j] += dp[i][j] * 2;
            if(j-a[i]>=0)dp[i+1][j] += dp[i][j-a[i]]; 
            dp[i+1][j] %= mod;
        }
    }
    cout << dp[n][s] << endl;


}