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 <cstdio>
#include <vector>
#include <iomanip>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
using namespace std;

int main() {
    int N;
    int S;
    cin >> N >> S;
    // vector <int> v(N);
    const long long MOD = 998244353;

    vector <long long> dp1(S+1);
    // dp[n][v] = dp[n-1][v-vx] + dp[n-1][v]

    dp1[0] = 1;
    int sum = 0;

    for (int i = 0; i < N; ++i) {
        int v;
        cin >> v;

        vector <long long> dp2(S+1);

        for (int val = 0; val <= sum; ++val) {
            dp2[val] += dp1[val]*2;
            dp2[val] %= MOD;

            if (val+v <= S) {
                dp2[val+v] += dp1[val];
                dp2[val+v] %= MOD;
            }
        }

        sum = min(sum+v, S);
        dp1 = dp2;
        // cin >> v[i];
    }

    cout << dp1[S] << endl;

    return 0;
}