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: Time Limit Exceeded)

import numpy as np
import sys
sys.setrecursionlimit(10**7)
input = sys.stdin.readline
n,k=map(int,input().split())
a=np.array(list(map(int,input().split())))
dp=np.full(k+1,False)
calted=np.full(k+1,False)
# dp[i]:残り石i個で自分のターンになった時、勝てるならTrue
# dp[0]=False,dp[a]=True,求めるのはdp[k]
mi=np.min(a)
ma=np.max(a)
dp[a]=True
calted[a]=True
# dp[i]={dp[i-a]の中にFalseがあればTrue、すべてTrueならFalse}
for i in range(mi+1,k+1):
      if calted[i]:
            continue
      il=i-a
      li=[j for j in il if j>=0]
      f=np.prod(dp[il])
      if f == 0:
            dp[i]=True
      else:
            dp[i]=False
print('First' if dp[k] else 'Second')
#print(dp)