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 <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
#include <random>
#include <stack>
#include <set>
#include <unordered_set>

#define bug(x) cout<<"zdongdebug1: "<<x<<endl;
#define bug2(x, y) cout<<"zdongdebug2: "<<x<<" "<<y<<endl;
#define bug3(x, y, z) cout<<"zdongdebug3: "<<x<<" "<<y<<" "<<z<<endl;
using namespace std;
typedef long long ll;
const int maxn = 3005;
const int mod = 998244353;


int f[maxn][maxn];
int a[maxn];

int main() {
#ifdef suiyuan2009
  freopen("/Users/suiyuan2009/CLionProjects/icpc/input.txt", "r", stdin);
  //freopen("/Users/suiyuan2009/CLionProjects/icpc/output.txt", "w", stdout);
#endif
  int n,m;
  cin>>n>>m;
  for(int i=1;i<=n;i++)cin>>a[i];
  f[0][0] = 1;
  for(int i=1;i<=n;i++)f[i][0] = f[i-1][0] * 2 %mod;
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++){
      f[i][j] = f[i-1][j] * 2 % mod;
      if(j>=a[i])f[i][j] = (f[i][j]+f[i-1][j-a[i]])%mod;
    }
  cout<<f[n][m]<<endl;
  return 0;
}