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>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define repn(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, a) for(int i = 0; i < (a); i++)
#define all(v) v.begin(), v.end() 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define endl '\n'

int dp[100005];

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
	int n, k;
	cin >> n >> k;
	int a[n + 1];
	repn(i, 1, n + 1){
		cin >> a[i];
	}
	repn(i, 1, k + 1){
		repn(j, 1, n + 1){
			if(i >= a[j] && !dp[i - a[j]]){
				dp[i] = 1;
			}
		}
	}
	cout << (dp[k] ? "First" : "Second") << endl;
	return 0;
}
/*
Things to look out for:
	- Integer overflows
	- Array bounds
	- Special cases
Be careful!
*/