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 pair<int, int> pii;
typedef pair<int, pair<int, int> > piii;
#define pb push_back
#define debug(x) cout << x << endl
#define fastio ios_base::sync_with_stdio(0), cin.tie(0)
#define PI acos(-1)
#define all(c) c.begin(), c.end()
#define SET(x, y) memset((x), y, sizeof(x))
const int MOD = 1e9 + 7;
const int INF = 1e9;
const LL INF64 = 1e18;

const int N = 1e5 + 5;

int n, k;
int dp[N][2];
int arr[105];

int main() {
	scanf("%d %d", &n, &k);
	dp[0][0] = 1;
	dp[0][1] = 0;
	for (int i = 0; i < n; i++) {
		scanf("%d", arr + i);
	}
	for (int i = 1; i <= k; i++) {
		dp[i][0] = 1;
		dp[i][1] = 0;
		for (int j = 0; j < n; j++) {
 			if (i < arr[j]) {
				dp[i][0] = min(1, dp[i][0]);
				dp[i][1] = max(0, dp[i][1]);
 			}
			else {
				dp[i][0] = min(dp[i][0], dp[i - arr[j]][1]);
				dp[i][1] = max(dp[i][1], dp[i - arr[j]][0]);
			}
		}
		// cout << dp[i][0] << dp[i][1] << endl;
	}
	if (dp[k][0] == 0)
		puts("First");
	else puts("Second"); 
	return 0;
}