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;
int main(){
int N,K;
  cin>>N>>K;
  vector<bool>dp(K+1);
  vector<int>a(N);
  for(int i=0;i<N;i++)
    cin>>a.at(i);
  for(int i=0;i<a.at(0);i++)
    dp[i]=false;
  for(int i=0;i<N;i++)
    dp[a.at(i)]=true;
  set<int>S;
  for(int i=0;i<N;i++)
    S.insert(a.at(i));
  for(int i=a.at(0)+1;i<=K;i++)
    if(!S.count(i)){
    bool x=false;
      for(int j=0;j<N;j++){
      if(0<=i-a.at(j)){
       if(!dp[i-a.at(j)]){x=true; break;}
      
      }
      }
         dp[i]=x;
    
    }if(dp[K])
      cout<<"First"<<endl;
  else
    cout<<"Second"<<endl;
    
    
    
  
   return 0;
}