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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n)    for(long long i = 0; i < (n) ; ++i)
#define orep(i,n)   for(long long i = 1; i <= (n); ++i)   // one rep
#define drep(i,n)   for(long long i = (n)-1; i >= 0; --i) // down rep
#define srep(i,s,t) for(long long i = (s); i < (t); ++i)  // set rep
#define rng(x)  (x).begin(),(x).end()

  // range
#define rrng(a) (a).rbegin(),(a).rend() // reverse range
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define ru(x,y) (((x)+(y)-1)/(y)) // round up
#define fi first
#define se second
#define eb emplace_back
#define fcout cout << fixed << setprecision(15)


using namespace std;
using ll  = long long;
using P   = pair<int,int>;
using vi  = vector<int>;
using vvi = vector<vi>;
using vp  = vector<P>;
using vl  = vector<ll>;

template<typename T> void Yes(T flag) {
  cout << (flag ? "Yes" : "No") << endl;
}

template<typename T> void pv(vector<T> vec) {
  cout << "[";
  for(auto& v : vec) { cout << v << ","; }
  cout << "]" << endl;
}

template<>
void pv(vector<P> vec) {
  cout << "[";
  for(auto& v : vec) {
    cout << "<" << v.fi << ","<< v.se << ">" << ",";
  }
  cout << "]" << endl;
}

template<typename T> void pv2(vector<vector<T>> vec) {
  for(auto& v : vec) pv(v);
}

bool dp[100005];

int main() {
  int n,k; cin >> n >> k;
  vi a(n,0);
  rep(i,n) cin >> a.at(i);

  dp[0] = false;
  orep(i,k) {
    rep(j,n) if(i - a.at(j) >= 0 and !dp[i-a.at(j)]) dp[i] = true;
  }

  if(dp[k]) cout << "First" << endl;
  else cout << "Second" << endl;

  return 0;
}