stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <iostream>

using namespace std;





int main() {



	int judge[1000000];

	for(int k=0;k<1000000;k++)judge[k]=1;

	int n,sum;

	

	while(cin>>n){

		if(n==0||n==1){

			sum = 0;

		}else{

			sum = 1;

			for(int i=3;i<n+1;i+=2){

				if (judge[i]==1){

					sum++;

					for(int j=2;j<=n/i;j++)

						judge[i*j] = 0;

				}	

			}

		}

		

		cout << sum <<endl;

	}



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.090573787422808x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



int main(){

    vector<int> ns;

    int temp, max_n = 0;

    while(cin >> temp){

        ns.push_back(temp);

        max_n = max(temp, max_n);

    }

    max_n++;



    vector<bool> is_prime(max_n, true);

    vector<int> primes;

    for(ll i=2;i<max_n;i++){

        if(is_prime[i]){

            primes.push_back(i);

            for(ll j=i*i;j<max_n;j+=i){

                is_prime[j] = false;

            }

        }

    }



    for(int n: ns){

        int ans = upper_bound(primes.begin(), primes.end(), n) - primes.begin();

        cout << ans << endl;

    }



    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7319720892333643x):
#include <bits/stdc++.h>

using namespace std;

#define print printf("==================\n")

#define ll long long

#define pi acos(-1.0)

#define eps 1e-16

#define max3(a,b,c) max(a,max(b,c))

const ll INF = 1 << 30;

#define mod 1000000007

typedef pair<double, double  > payar;

typedef struct

{

    ll  x, y;

} point;

priority_queue<payar, vector<payar>,  greater<payar> > pq; ///accending

vector< pair<double,payar  > > vpp;

vector < payar >  vp;

int n, k;

int arr[100005];

int dp[1000005];

int rec(int x){

	if(dp[x] != -1)return dp[x];

	int check = 0;

	for(int i = 0; i < n; i++){

		if(x + arr[i] <= k){

			if(rec(x + arr[i]) == 0)check = 1;

		}

	}

	return dp[x] = check;

}

int main(){



cin>>n>>k;

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

for(int i = 0 ; i<=k ; i++)dp[i]=-1;

if(rec(0)==1){

    cout<<"First";

}

else{

    cout<<"Second";

}

return 0 ;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3078314118216374x):
#include<set>

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;



ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int dp[maxn];//0 必败态; 1必胜态; -1尚未确定;

int a[105];

int n, k;

int cnt = 0;



bool dfs(int now){

	if(~dp[now]) return dp[now];

	bool ret = 0;

	rep(i, 1, n){

		if(a[i] > now) break;

		if(!dfs(now - a[i])) ret = 1;

	}

	return dp[now] = ret;

}



int main()

{

	n = read(), k = read();

	rep(i, 1, n) a[i] = read();

	memset(dp, -1, sizeof dp);

	sort(a + 1, a + n + 1);

	dp[0] = 0;

	dfs(k);

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

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8979868917945x):
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

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

#define forn(i, a, b) for (int i = a; i <= b; i++)

#define boost ios_base::sync_with_stdio(false); cin.tie(0),cout.tie(0);

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = (1e9) + 1;

const int mod = 1e9 + 7;

const int maxn = 2e5 + 111;

     

using namespace std;



int n, q, a[maxn], dp[maxn], k, mn = inf;

                           

void solve(){

	cin >> n >> k;



    for (int i = 1; i <= n; i++){

    	cin >> a[i];

    	mn = min(mn, a[i]);

    }



    sort(a + 1, a + n + 1);



   	for (int i = mn; i <= k; i++){

   		for (int j = 1; j <= n; j++){

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

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

   					dp[i] = 1;

   			}

   			else

   				break;

   		}

   	}



   	if (dp[k] == 1)

   		cout << "First";

   	else

   		cout << "Second";

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.929751558288162x):
//#pragma GCC target ("avx2,sse2")

//#pragma GCC optimization ("Ofast")

//#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;



#define ordered_set tree <pair <ll, int>, null_type, less <pair <ll, int> >, rb_tree_tag, tree_order_statistics_node_update>

#define ll long long

#define ull unsigned long long

#define db double

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define fi first

#define se second

#define mp make_pair

#define up_b upper_bound

#define low_b lower_bound

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

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

#define endl "\n"



#define left fsdsdfoisf

#define sum dpsdfioppsf

#define assign xcvjlkdjfio

#define trie fksdfkjkfnjuiv

#define next sidlfjsfkl

#define merge sdfksdkfsldf



//#define int long long



using namespace std;



void dout() {

    cerr << endl;

}

template <typename Head, typename... Tail>

void dout(Head H, Tail... T) {

    cerr << H << ' ';

    dout(T...);

}



//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef pair <int, int> pii;



const int N = 107, K = 1e5 + 7;

int n, k, a[N], dp[K];



void solve(int tc) {

    // check for (int i = 0; i < n; j++)

    cin >> n >> k;

    for (int i = 1; i <= n; i++) {

        cin >> a[i];

    }

    for (int i = 1; i <= k; i++) {

        for (int j = 1; j <= n; j++) {

            if (a[j] > i) {

                continue;

            }

            dp[i] |= (1 - dp[i - a[j]]);

        }

    }

    if (dp[k]) {

        cout << "First";

    }   else {

        cout << "Second";

    }

}



signed main() {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

//    freopen("input.txt", "r", stdin);

//    freopen("output.txt", "w", stdout);

    int tc = 1;

//    cin >> tc;

    for (int i = 0; i < tc; i++) {

        solve(i);

//        cleanup();

    }

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7635553282899423x):
#include <bits/stdc++.h>



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

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

#define ford(i, n) for (int i = int(n)-1; i >= 0; --i)

#define pb push_back

#define eb emplace_back

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

#define endl '\n'

#define de(x) cout<<#x<<" = "<<(x)<<endl

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int n,k;

	cin>>n>>k;

	vector<int> a(n);

	for(auto& it:a) cin>>it;

	vector<int> sg(k+1);

	for(int i=0;i<=k;i++){

		vector<bool> flag(k+1);

		for(auto it:a){

			if(i-it>=0) flag[sg[i-it]]=1;

		}

		for(int j=0;j<=k;j++){

			if(flag[j]==0){

				sg[i]=j;

				break;

			}

		}

	}

	cout<<(sg[k]?"First":"Second");

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.9174037829212227x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define F first

#define S second

#define endl '\n'

#define mod 1e9 + 7

const ll maxN = 1e5 + 5;

const int INF = 0x3f3f3f3f;



int main() {

	Omar45

    int n, k; cin >> n >> k;

    int arr[n];

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

    bool dp[k+1] = {0};

    for (int s=0; s<=k; s++){

        for (int it: arr){

            if (s >= it && !dp[s-it])

                dp[s] = 1;

        }

    }

    string ans = dp[k] ? "First" : "Second";

    cout << ans << endl;

}
// Slower C++ solution:
#include<set>

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;



ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int dp[maxn];//0 必败态; 1必胜态; -1尚未确定;

int a[105];

int n, k;

int cnt = 0;



bool dfs(int now){

	if(~dp[now]) return dp[now];

	bool ret = 0;

	rep(i, 1, n){

		if(a[i] > now) break;

		if(!dfs(now - a[i])) ret = 1;

	}

	return dp[now] = ret;

}



int main()

{

	n = read(), k = read();

	rep(i, 1, n) a[i] = read();

	memset(dp, -1, sizeof dp);

	sort(a + 1, a + n + 1);

	dp[0] = 0;

	dfs(k);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4660939977351752x):
#include <bits/stdc++.h>

using namespace std;



#define Omar45 ios_base::sync_with_stdio(false); cin.tie(NULL);

#define ll long long

#define pb push_back

#define F first

#define S second

#define endl '\n'

#define mod 1e9 + 7

const ll maxN = 1e5 + 5;

const int INF = 0x3f3f3f3f;



int main() {

	Omar45

    int n, k; cin >> n >> k;

    int arr[n];

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

    bool dp[k+1] = {0};

    for (int s=0; s<=k; s++){

        for (int it: arr){

            if (s >= it && !dp[s-it])

                dp[s] = 1;

        }

    }

    string ans = dp[k] ? "First" : "Second";

    cout << ans << endl;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.9406570516159842x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int N=1e5+5;

int main()

{

    int n,w;

    cin>>n>>w;

    int arr[n+2],dp[w+5];

    for(int i=0; i<n; i++)

        cin>>arr[i];

    memset(dp,0,sizeof dp);

    for(int i=0; i<=w; i++)

    {

        for(int j=0; j<n; j++)

        {

            if(i-arr[j]>=0&&dp[i-arr[j]]==0)

                dp[i]=1;

        }

    }

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

    return 0;

}
// Slower C++ solution:
#include<set>

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;



ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int dp[maxn];//0 必败态; 1必胜态; -1尚未确定;

int a[105];

int n, k;

int cnt = 0;



bool dfs(int now){

	if(~dp[now]) return dp[now];

	bool ret = 0;

	rep(i, 1, n){

		if(a[i] > now) break;

		if(!dfs(now - a[i])) ret = 1;

	}

	return dp[now] = ret;

}



int main()

{

	n = read(), k = read();

	rep(i, 1, n) a[i] = read();

	memset(dp, -1, sizeof dp);

	sort(a + 1, a + n + 1);

	dp[0] = 0;

	dfs(k);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4838740177626595x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int N=1e5+5;

int main()

{

    int n,w;

    cin>>n>>w;

    int arr[n+2],dp[w+5];

    for(int i=0; i<n; i++)

        cin>>arr[i];

    memset(dp,0,sizeof dp);

    for(int i=0; i<=w; i++)

    {

        for(int j=0; j<n; j++)

        {

            if(i-arr[j]>=0&&dp[i-arr[j]]==0)

                dp[i]=1;

        }

    }

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

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

const int MAX=1e5 + 10,MOD=1e9 + 7;

int N,K,A[105],mn=1e9;

int dp[MAX][2];

int f(int rem,int turn)

{

    if(rem<mn&&turn==1)return 0;

    if(rem<mn&&turn==0)return 1;

    if(dp[rem][turn]!=-1)return dp[rem][turn];

    for(int i=0;i<N;i++)

    {

        if(rem-A[i]>=0)

        if(f(rem-A[i],turn^1)==turn){

            return dp[rem][turn] = turn;

        }

    }

    return dp[rem][turn] = turn^1;

}

int main ()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);



    memset(dp,-1,sizeof dp);

    cin >> N >> K;

    for(int i=0;i<N;i++){cin>>A[i];mn = min(mn,A[i]);}

    if(f(K,0)==0){

        cout<<"First\n";

    }

    else cout<<"Second\n";

    //main();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.9581342242959763x):
#include <bits/stdc++.h>



using namespace std;

typedef long long ll;



const int N = 100007;



bool f[N];

int a[1007];

int main() {

    int n, k; cin >> n >> k;



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



    for(int i = 1; i <= k; ++i) {

        for(int j = 0; j < n; ++j) {

            if(i - a[j] >= 0) f[i] = f[i] or (!f[i - a[j]]);

        }

    }



    cout << (f[k]? "First" : "Second") << endl;

}