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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ffor(n) for(int i = 0; i < n; i++)
#define fffor(n) for(int j = 0; j < n; j++)
int INF = 1e9 + 7;

int main(void) {
  int n, k;
  cin >> n >> k;
  vector <int> v(n);
  ffor(n) cin >> v[i];
  sort(v.begin(), v.end());
  vector <int> dp(k + 1, 0); // dp[i] is 1 if winning, 2 if losing, 0 if unknown
  for(auto e: v) {
    dp[e] = 1;
  }
  for(int i = 1; i < v[0]; i++) dp[i] = 2;
  for(int i = 1; i <= k; i++) {
    if (dp[i] != 0) continue;
    bool can_force_loss = false;
    for (auto x: v) {
      if (i - x < 0) break;
      if (dp[i - x] == 2) can_force_loss = true;
    }
    if (can_force_loss) dp[i] = 1;
    else dp[i] = 2;
  }
  if (dp[k] == 1) cout << "First\n";
  else cout << "Second\n";
}