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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function Main(input) {
  input = input.split("\n");
  var N = parseInt(input[0].split(" ")[0]);
  var S = parseInt(input[0].split(" ")[1]);
  var A = input[1].split(" ").map(i => parseInt(i)).sort(function(a,b){return a-b});
  var mod = 998244353n;
  //var mod = 7n
  var memo = new Array(S);
  for(var i=0;i<S+1;i++){
    var memosub = [i,0n];
    memo[i]=memosub;
  }
  function modpow(a, n, mod) {
    res = 1n;
    while (n > 0) {
        if (n & 1n) res = res * a % mod;
        a = a * a % mod;
        n >>= 1n;
    }
    return res;
  }
  function modinv(a, mod) {
    return modpow(a, mod - 2n, mod);
  }
  var inv = modinv(2n,mod);
  var y = 1n;
  for(var i=0;i<N-1;i++){
    y = (y * 2n)%mod;
  }
  memo[0] = [0,1n];
  var ans = 0n;
  for(var i=0;i<N;i++){
    for(var j=memo.length-1;j>=0;j--){
      if(memo[j][0]+A[i]<=S&&j==0){
        memo[j+A[i]][1]=(memo[j+A[i]][1]+y)%mod;
      }else if(memo[j][0]+A[i]<=S){
        memo[j+A[i]][1]=(memo[j+A[i]][1]+(memo[j][1]*inv)%mod)%mod;
      }
    }
  }
  var ans = memo[memo.length-1][1].toString();
  console.log('%s', ans);
  return ans;
}

function debug(n){
  var input = document.getElementById("input"+n).value;
  var ans=Main(input);
  var result="WA";
  if(ans==document.getElementById("output"+n).value.split("\n")[0]){
    result="AC";
  }
  document.getElementById("result"+n).innerHTML=result;
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));