// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

#define inf (int)(1e9+7)

#define abs(x) (x >= 0 ? x : -(x))

#define ceil(a, b) a / b + !!(a % b)

template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }

ll pow(ll a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }



int main() {

    // 山の数は1つで, 取れる石の数が決まってる

    int n, k;

    cin >> n >> k;

    vector<int> a(n);

    rep(i, n) cin >> a[i];

    

    int dp[100001] = {0};

    rep3(i, 1, k + 1) {

        rep(j, sz(a)) {

            if (i - a[j] < 0) continue;

            dp[i] |= (dp[i - a[j]] == 0); // 1つでも遷移元に0があれば

        }

    }

    // rep(i, k + 1) cout << dp[i] << " "; cout << '\n';

    cout << (dp[k] == 0 ? "Second" : "First") << '\n';

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

const int inf = 1e9 + 7;

const ll INF = 1e18;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }

template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

const int mod = 10;

ll modpow(ll a, int b) { return b ? modpow(a * a % mod, b / 2) * (b % 2 ? a : 1) % mod : 1; }

template<class T> ostream& operator<<(ostream& os, const vector<T>& v) { rep(i, sz(v)) { if (i) os << " "; os << v[i]; } return os; }

template<class T, class U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

inline int add(int a, int b) { a += b; if (a >= mod) return a - mod; return a; }



void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);



    int T;

    // cin >> T;

    T = 1;



    while (T--) {

        solve();

    }

}







void solve() {

  int n, m;

  cin >> n >> m;

  vector<int> a(n);

  rep(i, n) cin >> a[i];

  sort(all(a));



  vector<int> dp(m + 1); // 0, 負け

  rep3(i, 1, m + 1) {

    rep(j, n) {

      if (i - a[j] < 0) break;

      if (dp[i - a[j]] == 0) {

        dp[i] = 1;

        break;

      }

    }

  }

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

  // rep(i, m + 1) cout << dp[i] << " "; cout << endl;

}


// Fastest known solution for this problem: