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: Accepted)

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

namespace AtCoderGrandContest010
{
    class ProgramA
    {
        private const string kYes = "YES";
        private const string kNo = "NO";
        static void Main(string[] args)
        {
            string szTotal = Console.ReadLine();
            int nTotal = int.Parse(szTotal);
            long[] arr = new long[nTotal];

            int nEven = 0;
            string szLine = Console.ReadLine();
            string[] szSplittedLine = szLine.Split(' ');
            
            for (int i = 0; i < szSplittedLine.Length; ++i)
            {
                long a = long.Parse(szSplittedLine[i]);
                if( a % 2 == 0) { ++nEven; }
                arr[i] = a;
            }

            int nOdd = nTotal - nEven;
            
            // 奇数が1個残る場合は無理
            if( nOdd % 2 != 0) { Console.WriteLine(kNo); return; }

            // 奇数ペアを消し、半分の偶数が生まれる
            nEven += nOdd / 2;
            // あ、もう必ず1個の偶数が残る
            Console.WriteLine(kYes);
        }
    }
}