#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define mp make_pair
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
typedef long long ll;
using namespace __gnu_pbds;

typedef tree<pair<ll,ll>,null_type,less<pair<ll,ll> >,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
/* ll t = 0;
ordered_multiset me;
...
me.insert({x, t++});
me.erase(me.lower_bound({x, 0}));
cout << me.order_of_key({x, 0}) << "\n";

///////////   t is insertion time
*/
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
typedef long double lld;

lld pi=3.14159265358;
ll powr(ll a,ll b)
{
    ll c=1;
    ll i;
    for(i=0;i<b;i++)
        c*=a;
    return c;
}
ll hell=1000000007;
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;
}
ll dp[100005];
ll n,i,j,k;
ll a[105];
set<ll> s;
ll solve(ll k)
{ 
    if(k<0)
        return -1;
    if(k==0)
        return 0;
    if(s.find(k)!=s.end())
        return 1;
    if(dp[k]!=-1)
        return dp[k];
    int f=0;
    for(i=1;i<=n;i++)
    {
        ll x=1+solve(k-a[i]);
        if(x%2)
            f=1;
    }
    dp[k]=f;
    return f;
}

int main()
{
    for(i=0;i<100005;i++)
            dp[i]=-1;
    cin>>n>>k;
    for(i=1;i<=n;i++)
        {cin>>a[i];
        s.insert(a[i]);
        }
    if(solve(k))
        cout<<"First";
    else
        cout<<"Second";
    return 0;
}




