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 LEN(x) (int)x.size()
#define ALL(x) x.begin(), x.end()

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;

const int MOD = 1'000'000'007;
const double EPS = 0.000'000'001;

inline int fcmp(double x, double y)
{ return (x < y + EPS ? (x < y - EPS ? -1 : 0) : 1);}

const int K = 1'000'10;
bool dp[K][2];

int main()
{
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    ios_base::sync_with_stdio(false);
    cout.precision(10);

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

    vi x(n);

    for(int& i : x) cin >> i;

    sort(ALL(x));

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

    for(int j = 0; j < k; j++)
    {
        for(int i : x)
        {
            if(i + j > k) break;

            dp[j + i][0] = min(dp[j][1], dp[j + i][0]);
            dp[j + i][1] = max(dp[j + i][1], dp[j][0]);
        }
    }

    string ans = !dp[k][0] ? "First" : "Second";

    cout << ans << endl;

    return 0;
}