#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, k, dp[N], a[N];
int main()
{
    cin >> n >> k;
    for(int i = 1; i <= n; i ++)
        cin >> a[i];
    dp[0] = 0;
    for(int i = 1; i <= k; i ++) //枚举当前取得的总和 
        for(int j = 1; j <= n; j ++) //枚举dp[i]是否可以通过dp[i-a[j]]所得1 
        {
            if(i - a[j] < 0) continue ; //越界判断 
            dp[i] |= (dp[i - a[j]] == 0); //状态转移方程，满足其一即可 
        }
    cout << (dp[k] ? "First" : "Second");
}

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, k, dp[N], a[N];
int main()
{
	cin >> n >> k;
	for(int i = 1; i <= n; i ++)
		cin >> a[i];
	dp[0] = 0;
	for(int i = 1; i <= k; i ++) //枚举当前取得的总和 
		for(int j = 1; j <= n; j ++) //枚举dp[i]是否可以通过dp[i-a[j]]所得1 
		{
			if(i - a[j] < 0) continue ; //越界判断 
			dp[i] |= (dp[i - a[j]] == 0); //状态转移方程，满足其一即可 
		}
	cout << (dp[k] ? "First" : "Second");
}