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.

Examples

Input

3 4
2 2 4


Output

6


Input

5 8
9 9 9 9 9


Output

0


Input

10 10
3 1 4 1 5 9 2 6 5 3


Output

3296
#include<bits/stdc++.h>
#define mod 998244353ll
#define ll long long
using namespace std;
ll mem[3003][3003], ara[3003], s, n;
ll dp(int pos, int sum)
{
    if(sum>s)return 0;
    if(pos==n)return (sum==s);
    if(mem[pos][sum]!=-1)return mem[pos][sum];
    return mem[pos][sum]= (2ll*dp(pos+1, sum)+dp(pos+1, sum+ara[pos]))%mod;
}
int main()
{
    scanf("%lld %lld", &n, &s);
    for(int i=0; i<n; i++)scanf("%lld", &ara[i]);
    memset(mem, -1, sizeof mem);
    printf("%lld\n", dp(0, 0));
    return 0;
}
