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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;


int main() {
  int n,k;
  cin>>n>>k;
  vector<int> a(n);
  for(int &x:a)
  {
    cin>>x;
  }
  /*
  vector<bool> dp(k + 1);
  for(int stones = 0; stones <= k; ++stones) 
  {
      for(int x : a)
       {
          if(stones >= x && !dp[stones-x]) 
          {
              dp[stones] = true;
          }
      }
  }

  puts(dp[k] ? "First" : "Second");
*/
  vector<bool> dp(k+1);
  for(int x: a)
  {
    for(int stones = x;stones<=k;++stones)
    {
      if(!dp[stones-x])
      {
        dp[stones] = 1;
      }
    }
  }
 puts(dp[k] ? "First" : "Second");
  /*
  if(helper(a,n,k,0)==0)
  {
    cout<<"first";
  }
  else
  {
    cout<<"second";
  }
*/
  return 0;
}