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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
#include<math.h>
//#include "Ants.h"
using namespace std;
typedef long long ll;
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
typedef pair < int, int> P;
/****Union-Ford-Tree***/
int par[int(0x7ffffff)];//Parent
int Rank[int(0x7ffffff)];//Deep_of_the_Tree

//int a[100] = { 4,2,3,1,5 }, m[100] ={3,2,2};
//int n = 0;
//const ll mod = 1000000007;


//n要素で初期化
void init(int n) {
	for (int i = 0; i < n; i++) {
		par[i] = i;
		Rank[i] = 0;
	}
}
//木の根探し
int find(int x) {
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find(par[x]);
	}
}
//xとyの属する集合を併合
void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if (x == y)return;

	if (Rank[x] < Rank[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (Rank[x] == Rank[y])Rank[x]++;
	}
}

bool same(int x, int y) {
	return find(x) == find(y);
}
/************************************************************************************/
int dx[4] = { 1,0/*,-1,0*/ }, dy[4] = { 0,1/*,0,-1*/ };
/**************/
int X, K, A[100];
bool win[int(1e5)+1];

int main()
{
	cin >> K>>X;
	for (int i = 0; i < K; i++)cin >> A[i];
	for (int j = 1; j <= X; j++) {
		win[j] = false;
		for (int i = 0; i < K; i++) {
			win[j] |= A[i] <= j && !win[j - A[i]];

		}
	}

	if (win[X])cout << "First" << endl;
	else cout << "Second"<<endl;

	
}