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)

// Sky's the limit :)
#include <bits/stdc++.h>
using namespace std;
#define int long long

const int mod = 998244353;

int power(int x, int y) {
    int res = 1;
    while(y) {
        if(y & 1) 
            res = res * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return res;
}

signed main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    
    int T = 1; 
    // cin >> T; 
    while (T--) {
        int n, s; 
        cin >> n >> s;
        int a[n];
        for(int i = 0; i < n; i++)
            cin >> a[i];

        int dp[n + 5][s + 5];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = power(2, n);
        int inv = power(2, mod - 2);
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= s; j++) {
                dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod;
                if (j + a[i] <= s) 
                    dp[i + 1][j + a[i]] = (dp[i + 1][j + a[i]] + dp[i][j] * inv) % mod;
            }
        }
     
        cout << dp[n][s] << '\n';
        
    }
    
    return 0;
}