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: Time Limit Exceeded)

#include <bits/stdc++.h>

const int INF = 1e9;
const int MOD = 998244353;

using namespace std;

typedef long long ll;

#define REP(i, n) for (int i = 0; i < int(n); i++)
#define REPD(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = a; i < int(b); i++)
#define FORD(i, a, b) for (int i = b - 1; i >= int(a); i--)
#define WRAP(y, x, h, w) (0 <= y && y < h && 0 <= x && x < w)

#define ALL(x) (x).begin(), (x).end()

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };

ll powmod(ll n, ll x)
{
    vector<ll> vll(32);
    vll[0] = n;
    FOR(i, 1, 32)
    {
        vll[i] = (vll[i - 1] * vll[i - 1]) % MOD;
    }
    ll ret = 1;
    int cnt = 0;
    while (x > 0) {
        if (x & 1) {
            ret = (ret * vll[cnt]) % MOD;
        }
        x >>= 1;
        cnt++;
    }
    return ret;
}

int main()
{
    int n, s;
    cin >> n >> s;
    vector<int> vi(n);
    REP(i, n)
    cin >> vi[i];
    sort(ALL(vi));
    map<pair<int, int>, int> mpii; //sum,use num
    vector<int> num(n, 0);
    mpii[{ 0, 0 }] = 1;
    REP(i, n)
    {
        map<pair<int, int>, int> nextmpii;
        for (auto itr = mpii.begin(); itr != mpii.end(); itr++) {
            if ((itr->first).first + vi[i] > s) {
                continue;
            }
            nextmpii[{ (itr->first).first, (itr->first).second }] += itr->second;
            if ((itr->first).first + vi[i] == s) {
                num[(itr->first).second] += itr->second;
            }
            if ((itr->first).first + vi[i] < s) {
                nextmpii[{ (itr->first).first + vi[i], (itr->first).second + 1 }] += itr->second;
            }
        }
        mpii = nextmpii;
    }
    ll ans = 0;
    REP(i, n)
    {
        ans = (ans + (powmod(2, n - i - 1) * num[i]) % MOD) % MOD;
    }
    cout << ans << endl;
}