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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#pragma region Imports
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<iomanip>
using namespace std;
#pragma endregion
#pragma region Macros
#pragma GCC optimize (2)
#define uns unsigned
#define out cout
#define in cin
#define au auto
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl "\n"
#define elif else if
#define sz(x) x.size()
#define fl(i, a, b) for(int (i)=a;(i)<(b);(i)++)
#define fli(i, a, b, c) for(int (i)=a;(i)<(b);(i)+=(c))
#define fr(i, x) for(int (i)=(x)-1;(i)>=0;(i)--)
#define fb(i, x) for(int (i)=0;(i)<(x);(i)++)
#define fa(i, x) for(auto& (i):(x))
#define ia(a, n) for(int qw=0;qw<(n);qw++)cin>>(a)[qw];
#define oa(a, n) for(int qw=0;qw<(n);qw++)cout<<(a)[qw]<<" ";
#define im(a, n, m) fb(i,(n))fb(j,(m))cin>>(a)[i][j];
#define om(a, n, m) fb(i,(n))fb(j,(m))cout<<(a)[i][j]<<" \n"[j==(m)-1];
#define eb(v, x) (v).emplace_back(x);
#define seb(s, x) (s).emplace(x);
#define pr(x, y) pair<x,y>
#define vr(x) vector<x>
#define mp(x, y) make_pair(x,y)
#define hm(x, y) map<x,y>
#define um(x, y) unordered_map<x,y>
#define stk(x) stack<x>
#define qu(x) queue<x>
#define dq(x) deque<x>
#define pq(x) priority_queue<x>
#define pqm(x) priority_queue<x,vector<x>,greater<(x)>>
#define mem(a, x) memset(a,(x),sizeof(a));
#define fi(x) (x).first
#define se(x) (x).second
#define ql(x) int tc;tc=(x);if(tc==-1)cin>>tc;while(tc--)
#define srt(x) sort((x).begin(),(x).end());
#define pops(y, x) auto (y)=(x).front();(x).pop();
#define ppc(x) __builtin_popcount((x))
#define all(x) (x).begin(),(x).end()
#define sieve(b, n) bool (b)[(n)];mem((b),true);int sq=int(sqrt(n))+1;(b)[0]=(b)[1]=false;fl(i,2,sq)if((b)[i])fli(j,i*i,n,i)(b)[j]=false;
#define dbg(x) cout<<#x<<" "<<(x)<<"\n";
#pragma endregion
#pragma region Typedefs and Consts
typedef long long int ll;
typedef unsigned long long int ull;
typedef string stg;
typedef float flo;
typedef char chr;
typedef vector<int> vi;
const ull mod = 1000000007;
const int imx = INT_MAX;
const int imn = INT_MIN;
const ll lmx = LLONG_MAX;
const ll lmn = LLONG_MIN;
#pragma endregion
#pragma region Functions
inline ull powm(ull x,ull y,bool b=false,ull m=mod){ if(b){ ull t=1;x%=m;while(y>0){ if(y%2)t=(t*x)%m;y=y>>1;x=(x*x)%m; }return t; }else
    { ull t=1;while(y>0){ if(y&2)t*=x;y=y>>1;x*=x; }return t;}}
inline ll gcd(ll a,ll b){a=abs(a);b=abs(b); while(b!=0){ ll c=a%b;a=b;b=c;}return a;}
#pragma endregion

//cout<<"Case #"+to_string(tc+1)+": "+to_string(z)<<endl;
int f(int a[],int n,int k,int p,hm(pr(int,int),int) d){
  if(k==0)
    return p^1;
  if(k<0)
    return p;
  if(d.find(mp(k,p))==d.end()){
    d[mp(k,p)]=p^1;
    fb(i,n)
      if(f(a,n,k-a[i],p^1,d)==p)
        return d[mp(k,p)]=p;
  }
  return d[mp(k,p)];
}
int main() {
  fio
  int n,k;
  in>>n>>k;
  int a[n];
  ia(a,n)
  hm(pr(int,int),int) m;
  out<<(f(a,n,k,0,m)==0?"First":"Second");
  return 0;
}