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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Text;
using System.Linq;
using System.Collections.Generic;


class Program {

	const long mod = 1000000007;
	//public static int[] max;

	static void Main(string[] args) {

		char[] cs = new char[] { ' ', ',' };
		int val = int.Parse(Console.ReadLine());
		int[] vals = parseAry(Console.ReadLine().Split(cs));
		//int val = int.Parse(Console.ReadLine());
		//string str = Console.ReadLine();
		//string[] strs = Console.ReadLine().Split(cs);

		//int[] vals = parseAry(Console.ReadLine().Split(cs));
		//int n = vals[0];
		//string res="";

		//int y = vals[0];
		//int x = vals[1];
		int guu = 0, ki = 0;
		for (int i = 0; i < val; i++) {
			if (( vals[i] & 1 ) == 1) { ki++; } else { guu++; }
		}

		string re="NO";
		if (( ki & 1 ) == 1) {
			//if (( (guu+(ki/2)) & 1 ) == 0) { re = "YES"; }
		} else {
			if (( (guu+(ki/2)) & 1 ) == 0) { re = "YES"; }
		
		}



			Console.WriteLine(re);

		
		return;
	}

		

	//　String[]  ----->  int[]
	static int[] parseAry(string[] str) {
		int[] nums = new int[str.Length];
		for (int i = 0; i < str.Length; i++) { nums[i] = int.Parse(str[i]); }
		return nums;
	}
	//　String[]  ----->  long[]
	static long[] parseAryl(string[] str) {
		long[] nums = new long[str.Length];
		for (int i = 0; i < str.Length; i++) { nums[i] = int.Parse(str[i]); }
		return nums;
	}



	//文字列を1文字ずつString[]に入れる
	static string[] strAry(String str) {
		string[] re = new string[str.Length];
		for (int i = 0; i < str.Length; i++) { re[i] = str[i].ToString(); }
		return re;
	}

}