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: Accepted)

	#include<bits/stdc++.h>
	using namespace std;
	typedef long long int ll;
	#define pb push_back
	#define mp make_pair
	#define sz(a) ll(a.size())
	#define trace(c,x) for(auto &x:c)
	#define F first
	#define S second		
	ll mod = 1e9 + 7;   //change acc to question

ll fastexpo(ll x,ll y)
{
	ll a=1;
	while(y>0)
	{
		if(y&1)a = (a*x)%mod;
		y/=2;
		x=(x*x)%mod;
	}	
	return a;
}

int memo[100005];
int a[103];
int n;

int dp(int i)
{
	// cout<<i<<endl;
	if(i==0)
		return 0;
	if(i<0)
		return 1;
	if(memo[i]!=-1)
		return memo[i];

	int val = 0;
	memo[i] = 0;
	for(int j=0;j<n;j++)
	{
		val = dp(i-a[j]);
		if(val==0)
			memo[i] = 1;
	}

	return memo[i];
}


int main()
{

	ios::sync_with_stdio(false);

	int k;
	cin>>n>>k;

	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<=k;i++)
		memo[i] = -1;

	if(dp(k)==1)
		cout<<"First"<<endl;
	else
		cout<<"Second"<<endl;
	return 0;

	}