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;
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)x.size()
#define D(x) cerr << #x << " = " << (x) << '\n'
using ll = long long;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	vector<vector<bool>> dp(2, vector<bool>(k + 1, true));
	for(int i = 0; i < n; ++ i) cin >> a[i];
	for(int i = 0; i <= k; ++ i) {
		for(auto x:a) {
			if(x > i) continue;
			dp[0][i] = dp[0][i] & dp[1][i - x];
			dp[1][i] = dp[1][i] & dp[0][i - x];
		}
		dp[0][i] = !dp[0][i];
		dp[1][i] = !dp[1][i];
	}
	cout << (dp[0][k] ? "First\n":"Second\n");
}