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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int en;
int on;
int N;

void input(){
	int i, t;
	en = 0;
	on = 0;
	scanf("%d", &N);
	for( i = 0; i < N; i++ ){
		scanf("%d", &t);
		if( t % 2 ){
			en++;
		}else{
			on++;
		}
	}
	return;
}

int main(){
	input();
	if( on % 2 == 0 || (on == 1 && en == 0) ){
		printf("YES\n");
	}else{
		printf("NO\n");
	}
	return 0;
}