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)

/* aman36 - Aman Raj */

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define fast_io ios :: sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define AfterDecimal(n) cout << fixed << setprecision(n);
#define pb push_back
#define ll long long
#define ld long double
#define all(a) a.begin(), a.end()
#define pll pair<ll, ll>
#define pii pair<int, int>
#define F first
#define S second
#define bits(x) __builtin_popcount(x)
#define fori(n)  for (ll i = 0; i < n; i++)
#define forj(n)  for (ll j = 0; j < n; j++)
#define fork(n)  for (ll k = 0; k < n; k++)
#define vll vector <ll>
#define mll map <ll, ll>
#define vpll vector <pll>
#define vi vector <int>
#define mem(x, v) memset(x, v, sizeof(x));
#define int ll
#define endl "\n"

// priority_queue<ll, vector<ll>, less<ll>>pq;     // Max Heap
// priority_queue<ll, vector<ll>, greater<ll>>pq;  // Min Heap

// template <typename T> using ordered_set =
//  tree<T, null_type, less<T>, rb_tree_tag, 
//  tree_order_statistics_node_update>;

#ifdef DEBUG
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template<typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cout << "# "<< name << " = " << arg1 << '\n';
}
template<typename Arg1, typename... Args>
void __f(const char* names, Arg1 && arg1, Args &&... args){
        const char* comma = strchr(names + 1 ,',');
        cout << "# ";
        cout.write(names, comma - names)<< " = " << arg1 << " | ";
        __f(comma + 1, args...);
}
#else
#define debug(...) 36
#endif

ll power(ll x, ll y, ll p) { 
    ll res = 1;
    x = x % p;  
    while (y > 0) {
        if (y & 1) 
            res = (res * x) % p; 
        y = y >> 1;
        x = (x * x) % p; 
    }
    return res;
}

bool isPrime(int n) 
{
    if (n <= 1) return false; 
    if (n <= 3) return true; 
    if (n % 2 == 0 || n % 3 == 0) return false; 
    for (int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) return false; 
    return true; 
}

const ll INF = (ll)(8e18);
const ll NEGINF = (ll)(-6e18);
const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const int M = 4000 + 5;
const int N = 1e5 + 5;
const int maxn = 1e5 + 5;

void solve() {
    int n, k;
    cin >> n >> k;

    int a[n];
    fori(n) cin >> a[i];

    int dp[k+1];
    mem(dp, 0);

    for(int i=1; i<=k; i++) {
        for(int j=0; j<n; j++) {
            if(i - a[j] >= 0 && dp[i - a[j]] == 0) dp[i] = 1;
        }
    }

    cout << (dp[k] ? "First" : "Second") << endl;
}

signed main() {
    fast_io;
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif

    int tc = 1;
    // cin >> tc;

    for(int i=1; i<=tc; i++) {
        // cout << "Case #" << i << ": " ;
        solve();
    }

    #ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
    #endif

    return 0;
}

// read the question correctly
// read code carefully
// corner cases // int vs ll // cin vs scanf // clear structures // maths
// check if max, min set to INT_MAX instead of INF
// check constraint