Problem:
Score : 100 points

Problem Statement

There is a set A = \{ a_1, a_2, \ldots, a_N \} consisting of N positive integers.
Taro and Jiro will play the following game against each other.

Initially, we have a pile consisting of K stones.
The two players perform the following operation alternately, starting from Taro:

Choose an element x in A, and remove exactly x stones from the pile.

A player loses when he becomes unable to play.
Assuming that both players play optimally, determine the winner.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq K \leq 10^5

1 \leq a_1 < a_2 < \cdots < a_N \leq K

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \ldots a_N

Output

If Taro will win, print First; if Jiro will win, print Second.

Sample Input 1

2 4
2 3

Sample Output 1

First

If Taro removes three stones, Jiro cannot make a move.
Thus, Taro wins.

Sample Input 2

2 5
2 3

Sample Output 2

Second

Whatever Taro does in his operation, Jiro wins, as follows:

If Taro removes two stones, Jiro can remove three stones to make Taro unable to make a move.

If Taro removes three stones, Jiro can remove two stones to make Taro unable to make a move.

Sample Input 3

2 7
2 3

Sample Output 3

First

Taro should remove two stones. Then, whatever Jiro does in his operation, Taro wins, as follows:

If Jiro removes two stones, Taro can remove three stones to make Jiro unable to make a move.

If Jiro removes three stones, Taro can remove two stones to make Jiro unable to make a move.

Sample Input 4

3 20
1 2 3

Sample Output 4

Second

Sample Input 5

3 21
1 2 3

Sample Output 5

First

Sample Input 6

1 100000
1

Sample Output 6

Second

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define ll int
#define pb push_back
#define mp make_pair
#define ld long double
#define sz(a) (ll)(a).size()

typedef tree<pair<ll,ll> ,null_type,less<pair<ll,ll> >,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//K-th smallest
//cout << k << "kth smallest: " << *A.find_by_order(k-1) << endl;
//NO OF ELEMENTS < X
//cout << "No of elements less than " << X << " are " << A.order_of_key(X) << endl;

const int MAXN = 105;
const int MAXK = 100005;

ll n, k, a[MAXN], dp[MAXK];      // Grundy number of val

ll rec(ll val)
{
	if(val < a[0]) return 0;
	if(dp[val] != -1) return dp[val];
	for(int i=0;i<n;i++)
		if(rec(val - a[i]) == 0) return dp[val] = 1;
	return dp[val] = 0;
}	

int main()
{
	std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

	cin >> n >> k;
	for(int i=0;i<n;i++) cin >> a[i];
	memset(dp, -1, sizeof(dp));
	if(rec(k)) cout << "First" << endl; 	   
    else cout << "Second" << endl;
    return 0;
}