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<ll, ll> pll;
typedef pair<int, int> pii;

#define endl "\n"

const ll INF = 1e15;
const int MOD = 1e9 + 7;
const double EPS = 1e-7;
const double PI = acos(-1.0);

const int N = 105;
const int K = 1e5 + 5;

int dp[K];
ll n, k, a[N];

int solve(int rem) {
    if (rem < a[0]) return 0;
    int& ret = dp[rem];
    if (~ret) return ret;
    ret = 0;
    for (int j = 0; j < n; ++j)
        if (rem >= a[j])
            ret |= !solve(rem - a[j]);
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    cin >> n >> k;
    for (int i = 0; i < n; ++i) cin >> a[i];
    memset(dp, -1, sizeof(dp));
    cout << (solve(k) ? "First" : "Second") << endl;
}