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 <ext/pb_ds/assoc_container.hpp>
    #include <ext/pb_ds/tree_policy.hpp>

    #define ll long long
    #define pb push_back
    #define mp make_pair
    #define f first
    #define s second
    #define sz(x) (int)x.size()
    #define all(x) x.begin(), x.end()

    using namespace std;
    using namespace __gnu_pbds;

    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
    template<class T1, class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
    template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
    template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
    template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
    template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
    template<class T1,class T2> ostream& operator<<(ostream& os, const gp_hash_table<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}

    int dp[100050];

    vector<int> a;
    int n, k;

    int nadji(int k)
    {
        //cout << "AAA " << k << endl;
        if(k < 0)
            return 1;
        if(dp[k]!=-1)
            return dp[k];




       //cout << "proveravam za k " << k << endl;


        for(int i =0; i<n; i++)
        {
             //cout << "proveravam za k-a " << k-a[i] << endl;

            if(nadji(k-a[i])==0)
            {
                //cout << "naso za " << k << endl;
                dp[k]=1;
                return 1;
            }
        }

        dp[k] = 0;
        return 0;


    }

    int main()
    {




        cin >> n>>k;
        int m=100050;
        for(int i=0; i <n; i++)
        {
            int pom;
            cin >> pom;
            a.push_back(pom);
            m=min(m, pom);
        }

        for(int i=0; i <=k; i++)
        {
            dp[i]=-1;
        }

        for(int i=0; i <m; i++)
        {
            dp[i]=0; //trenutni ne pobedjuje
        }

        int res = nadji(k);

        if(res==1)
            cout << "First" << endl;
        else
            cout << "Second" << endl;

        return 0;
    }