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)

/*
confirm 0LL and 1LL
confirm cornercases such as 0
confirm times of cin < 10^6
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using Pld = pair<ld, ld>;
using Vec = vector<ll>;
using VecP = vector<P>;
using VecB = vector<bool>;
using VecC = vector<char>;
using VecD = vector<ld>;
using VecS = vector<string>;
using Graph = vector<VecP>;
template <typename T>
using Vec1 = vector<T>;
template <typename T>
using Vec2 = vector<Vec1<T> >;
#define REP(i, m, n) for(ll i = (m); (i) < (n); ++(i))
#define REPN(i, m, n) for(ll i = (m); (i) <= (n); ++(i))
#define REPR(i, m, n) for(ll i = (m)-1; (i) >= (n); --(i))
#define REPNR(i, m, n) for(ll i = (m); (i) >= (n); --(i))
#define rep(i, n) REP(i, 0, n)
#define repn(i, n) REPN(i, 1, n)
#define repr(i, n) REPR(i, n, 0)
#define repnr(i, n) REPNR(i, n, 1)
#define all(s) (s).begin(), (s).end()
#define pb push_back
#define fs first
#define sc second
template <typename T>
bool chmax(T &a, const T b){if(a < b){a = b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T b){if(a > b){a = b; return true;} return false;}
template <typename T>
ll pow2(const T n){return (1LL << n);}
template <typename T>
void cosp(const T n){cout << n << ' ';}
void co(void){cout << '\n';}
template <typename T>
void co(const T n){cout << n << '\n';}
template <typename T1, typename T2>
void co(pair<T1, T2> p){cout << p.fs << ' ' << p.sc << '\n';}
template <typename T>
void co(const Vec1<T> &v){for(T i : v) cosp(i); co();}
template <typename T>
void co(initializer_list<T> v){for(T i : v) cosp(i); co();}
template <typename T>
void ce(const T n){cerr << n << endl;}
void sonic(){ios::sync_with_stdio(false); cin.tie(0);}
void setp(const ll n){cout << fixed << setprecision(n);}
constexpr int INF = 1e9+1;
constexpr ll LINF = 1e18+1;
// constexpr ll MOD = 1e9+7;
constexpr ll MOD = 998244353;
constexpr ld EPS = 1e-11;
const ld PI = acos(-1);

Vec fac, finv;

ll PowMod(ll a, ll n){
	if(n < 0) return PowMod(PowMod(a, -n), MOD-2);
	if(n == 0) return 1;
	if(n == 1) return a;
	if(n%2 == 0) return PowMod(a*a%MOD, n/2);
	return a*PowMod(a*a%MOD, n/2)%MOD;
}

ll inv(ll n){
	return PowMod(n, MOD - 2);
}

void init(ll n = 2e6){
	fac.resize(n+1);
	fac[0] = 1;
	repn(i, n) fac[i] = fac[i-1]*i%MOD;
	finv.resize(n+1);
	finv[n] = PowMod(fac[n], MOD-2);
	repr(i, n) finv[i] = finv[i+1]*(i+1)%MOD;
	finv[0] = 1;
}

ll combi(ll n, ll k){
	return fac[n]*finv[k]%MOD*finv[n-k]%MOD;
}

unordered_map<ll, ll> ump;
ll ans = 0;
ll N, S;

void f(ll m, ll s, ll c, ll t){
	// co({m, s, c, t});
	if(s == S){
		ans += PowMod(2, N - c) * t;
		ans %= MOD;
		return;
	}else if(s > S)
		return;

	if (m > 3000) return;

	rep(i, ump[m] + 1) f(m + 1, s + m * i, c + i, t * combi(ump[m], i) % MOD);
}

int main(void){
	init();
	cin >> N >> S;
	
	rep(i, N){
		ll a;
		cin >> a;
		++ump[a];
	}

	f(1, 0, 0, 1);
	co(ans);

	return 0;
}