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: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
const double pi = 3.14159265358979;

const int mod = 998244353;
int n, s, a[3001];

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n >> s;
	for(int i = 1; i <= n; i++) cin >> a[i];
	vector<long long> pow(n + 1);
	pow[0] = 1;
	for(int i = 1; i <= n; i++) pow[i] = (pow[i - 1] * 2) % mod;
	vector<vector<long long>> dp(n + 1, vector<long long>(s + 1));
	dp[0][0] = pow[n];
	for(int i = 1; i <= n; i++)
		for(int j = 0; j <= s; j++){
			if(j < a[i]) dp[i][j] = dp[i - 1][j];
			else dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - a[i]] / 2) % mod;
		}
	cout << dp[n][s] << "\n";
	return 0;
}