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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
#define zeroend __builtin_ctz
#define ones __builtin_popcount
#define ends "\n"
#define space ' '
typedef long long ll;
typedef pair<int, int> pii;

/*
void solve(int i, int j, string s){
	if(i==0)
		ans.insert(s2.substr(0, j)+s);
	else if(j==0)
		ans.insert(s1.substr(0, i)+s);
	else if(s1[i-1]==s2[j-1])
		solve(i-1, j-1, s1[i-1]+s);
	else{
		if(dp[i-1][j]<=dp[i][j-1])
			solve(i-1, j, s1[i-1]+s);
		if(dp[i-1][j]>=dp[i][j-1])
			solve(i, j-1, s2[j-1]+s);
	}
}*/


/*
	cin >> n;
	for(int i=0; i<n; ++i)
		cin >> a[i];

	for(int i=1; i<n; ++i){
		for(int j=0; j<i; ++j){
			if(a[j]>a[i]&&low[i]<high[j]+1)
				low[i]=high[j]+1;
			else if(a[j]<a[i]&&high[i]<low[j]+1)
				high[i]=low[j]+1;
		}
		ans=max({ans, low[i], high[i]});
	}

	cout << ans << ends;


*/

const int N = 100, N2=1e5, M = 1e9+7;

bool dp[N2+1];
//array<ll, 3> dp;
//set<string> ans;
//vector<int> low(N, 1), high(N, 1);
ll a[N];

ll n, x;

int main(){
	cin >> n >> x;
	for(int i=0; i<n; ++i)
		cin >> a[i];

	for(int i=1; i<=n; ++i){
		for(int j=a[i-1]; j<=x; ++j){
			dp[j]|=!dp[j-a[i-1]];
		}
	}

	cout << (dp[x]?"First":"Second") << ends;
}