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 gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define int long long int
#define si(x)   scanf("%d",&x)
#define sl(x)   scanf("%lld",&x)
#define ss(s)   scanf("%s",s)
#define pi(x)   printf("%d\n",x)
#define pl(x)   printf("%lld\n",x)
#define ps(s)   printf("%s\n",s)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define INF 100000000000
typedef pair<int, int>  pii;
typedef vector<int>     vi;
typedef vector<pii>     vpii;
typedef vector<vi>      vvi;
int mpow(int base, int exp);
void ipgraph(int m);
const int mod = 1000000007;
const int N = 2e5+5, M = N;
//=========================================


int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  
    // dp[i] = TRUE if the first player wins if there are i stones remaining
    int n, k;
    cin >> n >> k;

    vector<int> a(n);
    for(int i = 0; i < n; ++i)
        cin >> a[i];


    vector<bool> dp(k + 1);
    for(int stones = 0; stones <= k; ++stones){
        for(int x : a){
            if(stones >= x && !dp[stones - x])
                dp[stones] = true;
        }
    }

    puts(dp[k] ? "First" : "Second");

    return 0;
}