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)

//Resting_Prince
#include<bits/stdc++.h>
#define mod 1000000007
#define inf 1e18
#define PI 3.141592653589
#define fi first
#define se second 
#define sv(v) sort(all(v));reverse(all(v))
#define all(v) v.begin(),v.end()
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define repr(i,a,b) for(ll i=a;i>=b;i--)
#define pb push_back
#define mk make_pair
#define en '\n'
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double ld;

    ll n,k,i,j=inf,dp[100005]={0},a[2000];

ll mex(set<ll> s)
{
    ll x=0;
    while(s.find(x)!=s.end())x++;
    return x;
}

int main()
{
    cin>>n>>k;
    rep(i,0,n-1)cin>>a[i],j=min(a[i],j);
    rep(i,0,j-1)dp[i]=0;
    sort(a,a+n);
    rep(i,j,k)
    {
        set<ll> s;
        rep(i1,0,n-1)
        {
            if(a[i1]>i)break;
            s.insert(dp[i-a[i1]]);
        }
        dp[i]=mex(s);
    }
    if(!dp[k])cout<<"Second";
    else cout<<"First";
    return 0;
}