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<stdio.h>
#include<iostream>
#include<string.h>
#include <iomanip>
#include<math.h>
#include<queue>
#include<vector>
#define Max (int)1e5+10
#define MAX_INT (int)1e9+2
#define MOD ((int)1e9+7)
using namespace std;

int dp[Max];

void init(){
	int i;
	for(i=0;i<Max;i++)
		dp[i] = -1;
}

int canWin(int k,int min,int n,int a[]){
	if(k < min)
		return 0;
	if(dp[k] != -1)
		return dp[k];
	int status = 0,i;
	for(i=0;i<n;i++){
		if(canWin(k-a[i],min,n,a) == 0)
			status = 1;
	}
	dp[k] = status;
	return dp[k];
}

int main(){
	int n,k,a[Max],min=MAX_INT,i;

	init();
	scanf("%d%d",&n,&k);
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
		if(min > a[i])
			min = a[i];
	}

	if(canWin(k,min,n,a))
		printf("First\n");
	else
		printf("Second\n");

}










/*
Atcoder Dp G-Longest Path iterative solution

vector<int>vec[Max];
queue<int>q;

int dp[Max];
int vis[Max];

void init(){
	int i;
	for(i=0;i<Max;i++){
		dp[i] = 0;
		vis[i] = 0;
	}
}

int getMax(int a,int b){
	return (a<b)?b:a;
}

void solve(int x){
	int i,u;
	q.push(x);

	while(!q.empty()){
		u = q.front();
		q.pop();

		if(vis[u])
			continue;
		vis[u] = 1;
		
		for(i=0;i<vec[u].size();i++){
			dp[vec[u][i]] = getMax(dp[vec[u][i]],dp[u]+1);
			q.push(vec[u][i]);
		}
	}

}

int main(){
	int n,m,incoming[Max],x,y,i;
	for(i=0;i<Max;i++)
		incoming[i] = 0;
	
	init();

	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++){
		scanf("%d%d",&x,&y);
		vec[y].push_back(x);
		incoming[x]++;
	}
	
	for(i=1;i<=n;i++)
		solve(i);
	
	int max = 0;
	for(i=0;i<Max;i++)
		max = getMax(max,dp[i]);
	
	printf("%d\n",max);
	return 0;
}
*/