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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main{

    public static void main(String args[]){
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	long[] a=new long[n];
	for(int i=0;i<n;i++){
	    a[i]=sc.nextLong();
	}

	long sum=0;
	for(int i=0;i<n;i++){
	    sum+=a[i];
	}

	long consume=n*(n+1)/2;
	if(sum%consume!=0){
	    System.out.println("NO");
	    return;
	}

	long sump=sum/consume;
	long cnt=0;
	boolean isok=true;
	for(int i=0;i<n;i++){
	    //a[i-1]+(sump-p)=a[i]+(n-1)*p
	    long chil=(a[(i-1+n)%n]+sump-a[i]);
	    if(chil<0 || chil%n!=0) isok=false;
	    long p=chil/n;
	    if(p<0 || p>sump) isok=false;
	    cnt+=p;
	}
	if(isok && cnt==sump){
	    System.out.println("YES");
	}
	else{
	    System.out.println("NO");
	}
	return;
    }
    

    
}