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"
#include<random>
using namespace std;
typedef long long int lint;
typedef pair<lint, lint> plint;
typedef pair<double long, double long> pld;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(lint i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(lint i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2>& l, const pair<T1, T2>& r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2>& l, const pair<T1, T2>& r) { return make_pair(l.first - r.first, l.second - r.second); }
const lint MOD = 1e9 + 7, INF = 1e18;

lint dp[100001];
lint N, K, arr[100];

lint mex(lint curr) {
	set<lint> st;
	REP(i, N) {
		if (curr - arr[i] < 0) continue;
		if (dp[curr - arr[i]] == INF) continue;
		st.insert(dp[curr - arr[i]]);
	}
	REP(i, N + 1) {
		if (st.find(i) == st.end()) {
			return i;
		}
	}
}

int main() {
	cin >> N >> K;
	REP(i, N) {
		cin >> arr[i];
	}
	REP(i, K + 1) dp[i] = INF;
	REP(i, K + 1) {
		if (i < arr[0]) dp[i] = 0;
		else dp[i] = mex(i);
	}
	if (dp[K] == 0) cout << "Second" << endl;
	else cout << "First" << endl;
}