Problem:
Score : 300 points

Problem Statement

There are N integers written on a blackboard. The i-th integer is A_i.

Takahashi will repeatedly perform the following operation on these numbers:

Select a pair of integers, A_i and A_j, that have the same parity (that is, both are even or both are odd) and erase them.

Then, write a new integer on the blackboard that is equal to the sum of those integers, A_i+A_j.

Determine whether it is possible to have only one integer on the blackboard.

Constraints

2 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

A_i is an integer.

Input

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

N
A_1 A_2 … A_N

Output

If it is possible to have only one integer on the blackboard, print YES. Otherwise, print NO.

Sample Input 1

3
1 2 3

Sample Output 1

YES

It is possible to have only one integer on the blackboard, as follows:

Erase 1 and 3 from the blackboard, then write 4. Now, there are two integers on the blackboard: 2 and 4.

Erase 2 and 4 from the blackboard, then write 6. Now, there is only one integer on the blackboard: 6.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

NO

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

#include <cstdio>
#include <iostream>
using namespace std;
const int N=100005;
int n;
int a[N];
void readData(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
}
int gcd(int x,int y){
	if(x<y) swap(x,y);
	for(int z=x%y;z;x=y,y=z,z=x%y);
	return y;
}
void simulate(int who){
	static int sum[2],oddpos;
	bool all1flag=true;
	sum[0]=sum[1]=0;
	for(int i=1;i<=n;i++)
		if(a[i]!=1){
			all1flag=false;
			sum[a[i]&1]++;
			if(a[i]&1)
				oddpos=i;
		}
	if(sum[0]&1)
		throw who;
	if(sum[1]>1)
		throw who^1;
	if(sum[1]==0)
		throw who^1;
	a[oddpos]--;
	int g=-1;
	for(int i=1;i<=n;i++)
		if(a[i]!=1){
			if(g==-1)
				g=a[i];
			else
				g=gcd(g,a[i]);
		}
	for(int i=1;i<=n;i++)
		if(a[i]!=1)
			a[i]/=g;
	simulate(who^1);
}
int main(){
	readData();
	try{
		simulate(1);
	}
	catch(int e){
		puts(e?"First":"Second");
	}
	return 0;
}