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 Python (3.4.3) (judge verdict: Wrong Answer)


def main():
    n, k = map(int, input().split())
    a = tuple(map(int, input().split()))

    dp = [1] * (k + 1)
    dp[0] = 0  # 石がi個で手番が回ってきたら勝てる：1、負ける：0

    for i1 in range(1, k + 1):  # i1：手番が回ってきた時点の石の残数。残数１から始めてｋまでみる。
        if i1 < min(a):  # 「石の個数 < aの最小要素」だと取りようがないので負ける。
            dp[i1] = 0
            continue
        if i1 in a:  #  石の残数を０個にできるので勝てる。
            continue
        pattern1 = []
        pattern2 = []
        for i2 in range(n):
            if i1 - a[i2] > 0:
                pattern1.append((i1 - a[i2]) in a) # 石を取った残り個数がaにある＝次の手番で石を0にされる。
                for i3 in range(n):
                    if i1 - a[i2] - a[i3] >= 0:
                        pattern2.append(dp[i1 - a[i2] - a[i3]] == 0)
        if all(pattern1):
            dp[i1] = 0
        if any(pattern2):
            dp[i1] = 0
    first_te = []
    for j1 in range(n):
        first_te.append(dp[k - a[j1]] == 0)
    if any(first_te):
        print('First')
    else:
        print('Second')


if __name__ == '__main__':
    main()