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>                 // [PRIMES]               1777 ~2^10.80
using namespace std;                     //                       10333 ~2^13.33
using ll = long long;                    // seq 1 128 | factor   100333 ~2^16.61
using ld = long double;                  //   | grep -v ' .* '  1300111 ~2^20.31
using vl = vector<ll>;                   //                    10300777 ~2^23.30
using vvl = vector<vl>;                  //                   100400999 ~2^26.58
using pll = pair<ll,ll>;                 //                  1300400999 ~2^30.28
using vpll = vector<pll>;                //                 10200500333 ~2^33.25
using vb = vector<bool>;                 //                100200400777 ~2^36.54
using vs = vector<string>;               //               1200300700111 ~2^40.13
const ll oo = 0x3f3f3f3f3f3f3f3fLL;      //              10200300500777 ~2^43.21
const ld eps = 1e-9;                     //             100200300400777 ~2^46.51
#define sz(c) ll((c).size())             //            1200300400600999 ~2^50.09
#define all(c) begin(c),end(c)           //           10200300400600111 ~2^53.18
#define mp make_pair                     //          100200300400600333 ~2^56.48
#define mt make_tuple                    //         1200300400500800999 ~2^60.06
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define FOR(i,a,b) for(ll i=(a); i<(b); i++)
#define FORD(i,a,b) for(ll i=ll(b)-1;i>=(a);i--)
#define TR(X) ({if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);

	ll n, k; cin >> n >> k;
	vl as(n); FOR(i,0,n) cin >> as[i];

	vb dp(k+1);
	dp[0] = false;
	FOR(i,1,k+1) {
		bool next = false;
		for (ll a : as) {
			if (a <= i && !dp[i-a]) {
				next = true;
				break;
			}
		}
		dp[i] = next;
	}

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

	return 0;
}