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 Rust (1.15.1) (judge verdict: Accepted)

use std::cmp;
use std::collections::HashMap;
use std::collections::VecDeque;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

fn solve() {
    input! {
        n: usize,
        k: usize,
        a: [usize;n]
    }

    // dp[x][y] = x個目までのaを考慮し、石の数をyとした場合の勝者が先手なら0,後手なら1
    let mut dp = vec![2; k + 1];
    dp[0] = 1;

    for j in 1..k + 1 {
        let mut win = 1;
        for i in 1..n + 1 {
            let a_index = i - 1;
            if a[i - 1] == j {
                win = 0;
            } else if j > a[a_index] && dp[j - a[a_index]] == 1 {
                win = 0;
            }
        }
        dp[j] = win;
    }

    println!("{}", if dp[k] == 0 { "First" } else { "Second" });
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}

// 最大公約数
fn gcd(a: u64, b: u64) -> u64 {
    let mut aa: u64 = if a > b { a } else { b };
    let mut bb: u64 = if a > b { b } else { a };
    while bb != 0 {
        let tmp = bb;
        bb = aa % tmp;
        aa = tmp;
    }
    return aa;
}

// 最小公倍数
fn lcm(a: u64, b: u64) -> u64 {
    a / gcd(a, b) * b
}

// DFS,BFS
fn BFS() {
    let mut searchResult = vec![vec![MAX; 100]; 100];
    let mut q: VecDeque<(usize, usize)> = VecDeque::new();
    q.push_front((0, 0));
    searchResult[0][0] = 0;

    let moves = [(1, 0), (-1, 0), (0, 1), (0, -1)];

    while !q.is_empty() {
        let pos = q.pop_back().unwrap(); // これがpop_frontならstackになって深さ優先になる
        for m in moves.iter() {
            let dx: i64 = pos.0 as i64 + m.0;
            let dy: i64 = pos.1 as i64 + m.1;
            if dx >= 0
                && dx < 100
                && dy >= 0
                && dy < 100
                && searchResult[dx as usize][dy as usize] == MAX
            {
                q.push_front((dx as usize, dy as usize));
                searchResult[dx as usize][dy as usize] = searchResult[pos.0][pos.1] + 1;
            }
        }
    }
}

// 文字列操作

const LARGE_PRIME: u64 = 1_000_000_007;

fn power_mod(base: u64, exp: u64, modulo: u64) -> u64 {
    if exp == 0 {
        return 1;
    } else if exp & 1 == 0 {
        return (power_mod(base, exp / 2, modulo) % modulo).pow(2) % modulo;
    } else {
        return power_mod(base, exp - 1, modulo) % modulo * base % modulo;
    }
}

fn inverse_mod(element: u64, modulo: u64) -> u64 {
    // フェルマーの小定理からpが素数なら a^(p-1) = 1
    // よってa*a^(p-2) = 1 より a^(p-2)がaの逆元

    power_mod(element, modulo - 2, modulo) % modulo
}

fn combination_mod(m: u64, n: u64, modulo: u64) -> u64 {
    // m C n = m! / ( n! * (m - n)! )
    //       = m*(m-1)*(m-2)*...*(m-n+1)/n!
    //       = m*(m-1)*(m-2)*...*(m-n+1) * (n!)^-1

    let mut numerator = 1;
    for i in 0..n {
        numerator *= m - i;
        numerator %= modulo;
    }

    // n!
    let mut element = 1;
    for j in 1..(n + 1) {
        element *= j;
        element %= modulo;
    }

    numerator * inverse_mod(element, modulo) % modulo
}

fn topological_sort(n: usize, edges: Vec<(usize, usize)>) -> Vec<usize> {
    let mut result = vec![];
    let mut isolated = VecDeque::new();

    // to[x] = [a,b] x -> a, x -> b が存在
    let mut to = vec![vec![]; n + 1];

    // deg[x] = a -> x, b -> xのような辺の数
    let mut deg = vec![0; n + 1];
    for e in edges {
        to[e.0].push(e.1);
        deg[e.1] += 1;
    }

    for i in 1..n + 1 {
        if deg[i] == 0 {
            isolated.push_front(i);
        }
    }

    while isolated.len() != 0 {
        let next = isolated.pop_back().unwrap();
        result.push(next);
        for i in &to[next] {
            deg[*i] -= 1;
            if deg[*i] == 0 {
                isolated.push_front(*i);
            }
        }
    }

    result
}