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>
#define endl '\n'
#define ll long long int
#define loop(i,a,b)           for(ll i=a;i<=b;++i)
#define pb                    push_back
#define F                     first
#define S                     second
#define mp                    make_pair
#define clr(x)                x.clear()
#define MOD                   1000000007
#define itoc(c)               ((char)(((int)'0')+c))
#define vl                    vector<ll>
#define SZ(x)                 (x).size()
#define all(p)                p.begin(),p.end()
#define mid(s,e)              (s+(e-s)/2)
#define sv()                  ll t,n; scanf("%lld",&t);n=t; while(t--)
#define tcase()               ll t,n; cin>>t;n=t; while(t--)
#define iscn(num)             scanf("%d",&num);
#define scn(num)              scanf("%lld",&num);
#define scn2(num,num2)        scanf("%lld%lld",&num,&num2);
#define dbg(x)                cerr <<"( "<<#x << " = " << x << " )"<<endl
using namespace std;
const ll INF = 2e18 + 99;
typedef pair<ll,ll> Pair;
typedef vector<int> vec;;
void FAST_IO();
void ara_read(vec &v,ll n);
////////////////////////

int main()
{
	int n, k;
	scanf("%d%d",&n,&k);
	vec num;
	ara_read(num,n);
	// taking number of stones k, as dp dimension...
	vector< bool > dp(k+1);
	
	for(int stone=0; stone<= k; ++stone){
		for(auto x : num){
			if(stone>= x  && !dp[stone-x]){
				dp[stone]=true;
				break ;
			}
		}
	}
	// if dp[k] is true, that means first player is winner 
    puts((dp[k] ? "First" : "Second"));
    return 0;
}

void FAST_IO()
{
    // ios_base::sync_with_stdio(0);
    //cin.tie(NULL);
    //cout.tie(NULL);
    //cout.setf(ios::fixed);
    //cout.precision(2);

}
void ara_read(vec &v,ll n){
	v.resize(n);
	for(int & i : v){
		scanf("%d",&i);
	}
	
}