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>
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;

namespace mp = boost::multiprecision;

#define INF 100000000
#define YJ 1145141919
#define INF_INT_MAX 2147483647
#define INF_LL 9223372036854775
#define INF_LL_MAX 9223372036854775807
#define EPS 1e-10
#define MOD 1000000007
#define MOD9 998244353
#define Pi acos(-1)
#define LL long long
#define ULL unsigned long long
#define LD long double

#define int long long

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(a)  begin((a)), end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())

const int MAX_N = 105;
const int MAX_K = 100005;
int K;
int N;
int A[MAX_N];

int dp[MAX_K][2];

int solve(int size, int turn) 
{
  if(dp[size][turn] != -1) {
    return dp[size][turn];
  }

  int nextTurn = (turn+1)%2;

  REP(n,N) {
    if(size >= A[n]) {
      if(solve(size-A[n], nextTurn) == turn) {
        return dp[size][turn] = turn;
      }
    } else {
      break;
    }
  }

  return dp[size][turn] = nextTurn;
}

signed main()
{
  cin >> N >> K;
  REP(n,N) {
    cin >> A[n];
  }

  memset(dp, -1, sizeof(dp));
  if(solve(K, 0) == 0) {
    cout << "First" << endl;
  } else {
    cout << "Second" << endl;
  }

  return 0;
}