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;

#define int long long
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define fr(i,j,k) for(i = j; i < (k); i++)
#define all(x) x.begin(), x.end()
#define el '\n'
#define remax(a,b) a = max(a, b)
#define remin(a,b) a = min(a, b)
#define sz(x) int32_t(x.size())

typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpi;
// --------------------------------------------------------------------

const pii dxy[] = { {-1, 0}, {1, 0}, {0, 1}, {0, -1} };
const int mod = 998244353;
const int INF = 2e18;
const ld EPS = 1e-9;
const int NN = 3002;

int dp[NN][NN];
int a[NN];


void solve() {
      int i = 0, j = 0, k = 0, n = 0, m = 0; 
      int s;
      cin >> n >> s;
      fr(i, 0, n) {
            cin >> a[i];
      }
      int p2 = 1;
      dp[0][a[0]] = 1;
      fr(i, 1, n) {
                        p2 = (p2 * 2) % mod;

            fr(j, 0, s + 1) {
                  dp[i][j] = dp[i - 1][j] * 2;
                  dp[i][j] %= mod;
            }
            fr(j, 0, s + 1) {
                  if(a[i] + j <= s) {
                        int& v = dp[i][a[i] + j];
                        v = (v + dp[i - 1][j]) % mod;
                  }
            }
            int& v = dp[i][a[i]];
            v = (v + p2) % mod;
      }
      cout << dp[n - 1][s];
}

int32_t main() {
      ios::sync_with_stdio(false); cin.tie(0); 
      int T = 1; 
      // cin >> T;
      for(int tc = 1; tc <= T; tc++) {
            solve();
       }
      return 0;
}