Problem:
Score : 1000 points

Problem Statement

There are N integers written on a blackboard. The i-th integer is A_i, and the greatest common divisor of these integers is 1.

Takahashi and Aoki will play a game using these integers. In this game, starting from Takahashi the two player alternately perform the following operation:

Select one integer on the blackboard that is not less than 2, and subtract 1 from the integer.

Then, divide all the integers on the black board by g, where g is the greatest common divisor of the integers written on the blackboard.

The player who is left with only 1s on the blackboard and thus cannot perform the operation, loses the game.
Assuming that both players play optimally, determine the winner of the game.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

The greatest common divisor of the integers from A_1 through A_N is 1.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If Takahashi will win, print First. If Aoki will win, print Second.

Sample Input 1

3
3 6 7

Sample Output 1

First

Takahashi, the first player, can win as follows:

Takahashi subtracts 1 from 7. Then, the integers become: (1,2,2).

Aoki subtracts 1 from 2. Then, the integers become: (1,1,2).

Takahashi subtracts 1 from 2. Then, the integers become: (1,1,1).

Sample Input 2

4
1 2 4 8

Sample Output 2

First

Sample Input 3

5
7 8 8 8 8

Sample Output 3

Second

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define sqr(x) ((x)*(x))
#define mp make_pair
#define uint unsigned
#define PI pair<int,int>
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline int read(){
	int x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(int a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void writeln(int a){
    if(a<0){
    	a=-a; putchar('-');
	}
	write(a); puts("");
}
inline int rnd(int x){
	return rand()%x;
}
inline ull rnd(){
	return ((ull)rand()<<30^rand())<<4|rand()%4;
}
const int N=100005;
int n,a[N];
bool solve(){
	int sum=0;
	for(int i=1;i<=n;i++)if(a[i]%2==0)sum++;
	if(sum&1)return 1; else if(sum<n-1)return 0;
	for(int i=1;i<=n;i++)if(a[i]&1)a[i]--;
	int t=0;
	for(int i=1;i<=n;i++)t=__gcd(t,a[i]);
	for(int i=1;i<=n;i++)a[i]/=t;
	return solve()^1;
}
int main(){
	n=read(); 
	for(int i=1;i<=n;i++)a[i]=read();
	puts(n==1||solve()?"First":"Second");
}
/*
2 1
2 6
3 4
3 5

(2,3) (1,6) (1,4) (1,5) (
(1,2)-(1连出的点，2连出的点）
*/