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;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ll long long int
#define int ll
//#define slld(t) scanf("%lld",&t)
//#define sd(t) scanf("%d",&t)
//#define pd(t) printf("%d\n",t)
//#define plld(t) printf("%lld\n",t)
#define vec vector < int >
#define vecp vector < pair < int , int > >
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define debug(x)   cerr<<#x<<": "<<x<<endl
#define pii pair < int , int >
#define pll pair < ll , ll > 
#define debug(x)   cerr<<#x<<": "<<x<<endl
#define mod  1000000007
#define N 200005
#define mod1 1000000007
#define mod2 998244353
#define bitcnt __builtin_popcount
#define PI acos(-1)
//#define endl "\n"

const int inf = (int)1e18 + 10;

bool dp[100005] = {0};
int a[103];
signed main(){
  fast;
 

 int n , k;
 cin >> n >> k;
 for(int i = 0 ; i < n ; i++)
 	cin >> a[i];
 
 for(int i = 1 ; i<= k ; i++){
 	
 		bool d = 1;
 	   for(int j = 0 ; j < n ; j++){
 	   	    if(i - a[j] > 0)
                d &= dp[i-a[j]];
            else if(i - a[j] == 0)
                d = 0;

 	   }

 	   dp[i] = 1 - d;
   // cout << i << " "<< dp[i] << endl;
 }


 if(dp[k] == 1)
 	cout << "First\n";
 else
 	cout << "Second\n";

}