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;
class Scanner //標準入力
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int Int() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = Int() + add;
        }
        return Array;
    }

    public long Long() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = Long() + add;
        }
        return Array;
    }
}
class Program {
    static Scanner cin = new Scanner();
    static void Main(string[] args) {
        int N = cin.Int();
        var A = cin.ArrayInt(N);
        int odd = 0;
        for (int i = 0; i < N; i++) {
            if (A[i] % 2 != 0) odd++;
        }

        Console.WriteLine(odd % 2 != 0 ? "NO" : "YES");
        Console.ReadLine();
    }
}