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>
#include<stdio.h>
using namespace std;
typedef long long int ld;
typedef unsigned long long int lu;
typedef vector< int > vi;
typedef vector < ld > vd;
typedef set <ld> sd;
typedef vector< vi > vvi;
typedef vector< vd > vvd;
typedef pair< int,int > ii;
typedef pair< ld,ld > id;
#define sz(a) ld((a).size())
#define test(t) ll t;cin>>t;while(t--)
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(__typeof((c).begin()) i = (c).begin(); i != (c).end(); i++)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define Vi vector
#define pr pair
#define mpr make_pair
#define fi first
#define se second
#define isnum(a) (a>=48 && a<=57)
#define isS(a) (a>='a' && a<='z')
#define isU(a) (a>='A' && a<='Z')
#define toS(a) (isU(a)?a+32:a)
#define toU(a) (isS(a)?a-32:a)
#define toC(a) a&15

set <ld> s[100005];
ld d[100005];

vector<char> prime (100000+1, true);
void sieve(ld n)
{	
	//O(nloglogn) time
	prime[0] = prime[1] = false;
	for (ld i=2; i<=n; ++i)
	{
		if (prime[i])
		{
			s[i].insert(i);			
			for (ld j=2*i; j<=n; j+=i)
			{
				prime[j] = false;
				s[j].insert(i);
			}
		}
	}
}
#define forn(i,e) for(int i = 0; i < e; i++)
#define rep(i,a,b) for(i=a;i<b;i++)
#define REP(i,a,b) for(i=a;i<=b;i++)
#define reprev(i,a,b) for(i=a;i>=b;i--)

#define leadzero(a) __builtin_clz(a) //count leading zeros
#define trailzero(a) __builtin_ctz(a) //count trailing zeros
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln cout<<"\n"
#define sp cout<<" "
#define maxe(a) *max_element(a.begin(),a.end())
#define maxi(a) max_element(a.begin(),a.end())-a.begin()
#define mine(a) *min_element(a.begin(),a.end())
#define mini(a) min_element(a.begin(),a.end())-a.begin()
ld gcd(ld a, ld b) { if(b == 0) return a; return gcd(b, a % b); }
ld fastpowMOD(ld a, ld p,ld MOD){ if(p==0) return 1; ld z = fastpowMOD(a,p/2,MOD); z = (z*z)%MOD; if(p%2) z = (z*a)%MOD; return z; }
ld fastpow(ld a, ld p){ if(p==0) return 1; ld z = fastpow(a,p/2); z = (z*z); if(p%2) z = (z*a); return z; }
ld a,b;

ld search1(ld a[], ld f,ld n)
{
	ld left=1;
	ld right=n+1;
	while(right-left>1)
	{
		ld mid = left + (right-left)/2;
		if(a[mid]<f)
		{
			left=mid;
		}
		else right=mid;
	}
	return a[left];
}

int main(){
	ld n,k,i,j;
	cin >> n >> k;
	ld ar[n];
	rep(i,0,n)cin >> ar[i];
	ld dp[k+1];
	memset(dp,0,sizeof(dp));
	dp[0]=-1;
	REP(i,1,k)
	{
		dp[i] = -1;
		rep(j,0,n)
		{
			if(ar[j]>i)break;
			if(dp[i-ar[j]]==-1){dp[i]=1;break;}
		}
	}
	if(dp[k]==-1)cout << "Second" << endl;
	else if(dp[k]==1)cout << "First" << endl;
	return 0;
}