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: Accepted)

#include <stdio.h>

int main(void) {
	int gusu = 0, kisu = 0;
	int N, A, i;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &A) != 1) return 1;
		if (A % 2 == 0) gusu++; else kisu++;
	}
	puts(kisu % 2 == 0 || (gusu == 0 && kisu == 1) ? "YES" : "NO");
	return 0;
}

/*
gusu + gusu = gusu
kisu + kisu = gusu
gusu + kisu = kisu <- tukaenai
kisu + gusu = kisu <- tukaenai

偶奇が等しい数を選択する->奇数が奇数個だと1個取り残される
偶数はいくらでもたせる
奇数が1個 -> 自明にOK (2<=Nなので無いけど)
奇数が3個 -> 偶数1個と奇数1個に鳴るのでNG
*/