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"

using ll = long long;

const int MOD = 1000000007;

const int size = 100005;

void Break()
{
#if _MSC_VER
	static int x = 0;
	++x;
	switch (x)
	{
		__debugbreak();
	}
#endif
}

bool dp[size];

int main()
{
#if _MSC_VER
	freopen("input.txt", "r", stdin);
#endif
	int n, k;
	std::cin >> n >> k;
	std::vector<int> v(n);
	for (int& x : v)
	{
		std::cin >> x;
	}
	for (int i = 0; i <= k; ++i)
	{
		for (int x : v)
		{
			if (i >= x && !dp[i - x])
			{
				dp[i] = true;
			}
		}
	}
	if (dp[k])
	{
		std::cout << "First\n";
	}
	else
	{
		std::cout << "Second\n";
	}
	return 0;
}