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;

#define pb push_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define endl "\n"

#define aim_cpo do{ios::sync_with_stdio(false);\
cin.tie(nullptr);cout.precision(10);cout<<fixed;}while(0)

#define LOCAL_INPUT do{FILE *stream1;\
stream1=freopen("/Users/aim_cpo/CLionProjects/competitive/in.txt","r",stdin);\
if(stream1==nullptr)return 0;}while(0)

#define LOCAL_OUTPUT do{FILE *stream2;\
stream2 = freopen("out.txt", "w", stdout);\
if (stream2 == nullptr) return 0;}while(0)

#ifdef LOCAL_DEFINE
#define show(x) cerr << #x << " = " << x << endl
#else
#define show(x)
#endif

typedef long long ll;
typedef unsigned long long ull;

constexpr const int INT_INF=0x3f3f3f3f; // 1061109567
constexpr const ll LL_INF=0x3f3f3f3f3f3f3f3f; // 4557430888798830399

template <typename T> bool chmin(T &a, T b) {
    if (a > b) {a = b;return true;}return false;
}

template <typename T> bool chmax(T &a, T b) {
    if (a < b) {a = b;return true;}return false;
}

// INT
#define GCD(a, b) __gcd(a, b)
template <typename T> T LCM(T a, T b) {return a / GCD(a, b) * b;}
template <typename T> T EXTGCD(T a, T b, T& x, T& y) {
    T d = a;
    if (b != 0) {d=EXTGCD(b,a%b,y,x);y-=(a/b)*x;}
    else x=1,y=0;
    return d;
}
template <typename T> bool is_prime(T a) {
    for(int i=2;i*i<=a;i++)if(a%i==0)return 1;
    return 0;
}

// MOD
const ll MOD = (ll)1e9 + 7;
#define add(a, b) ((a % MOD) + (b % MOD)) % MOD
#define mul(a, b) ((a % MOD) * (b % MOD)) % MOD
#define sub(a, b) ((a % MOD) + MOD - (b % MOD)) % MOD
template <typename T> T mod_inverse(T a, T mod, bool prime){ // if mod is prime, "prime" is true.
    if(prime){
        T tmp=mod-2,now=a,res=1;while(tmp){if(tmp&1)res=mul(res,now);now=mul(now,now);tmp>>=1;}
        return res;
    }else{T x,y;EXTGCD(a,mod,x,y);return (mod+x%mod)%mod;}
}
#define divide(a, b) ((a % MOD) * (mod_inverse(b, MOD, true))) % MOD

//LLの数値をつかう時は最後にLをつける癖をつけよう
//
//  ┏┓┏┓　┓　┏┓
//  ┏┛┃┃　┃　┗┫
//  ┗┛┗┛　┻　┗┛
//
//   謹┃賀┃新┃年┃
//   ━┛━┛━┛━┛
//WWWWWWWWWWWWWWWWWWWWWW

int n, m, w, k;
int a[101];
bool dp[2][100001];
bool past[2][100001];

bool memo(int who, int num) {
    if (past[who][num]) return dp[who][num];
    bool win = false;
    for (int i = 0; i < n; i++) {
        if (num >= a[i]) {
            if (memo(who ^ 1, num - a[i]) == who) win = true;
        }
    }
    if (win) {
        past[who][num] = true;
        dp[who][num] = (bool)who;
    } else {
        past[who][num] = true;
        dp[who][num] = (bool)(who ^ 1);
    }
    return dp[who][num];
}

int main() {
    aim_cpo; // インタラクティブのときは消すように.
#ifdef LOCAL_DEFINE
    LOCAL_INPUT; // インタラクティブのときは消すように.
//    LOCAL_OUTPUT; // ファイルに出力したいときのみ
    show(MOD);
#endif
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    if (memo(0, k)) {
        cout << "Second" << endl;
    } else {
        cout << "First" << endl;
    }
#ifdef  LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;
#endif
    return 0;
}