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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.IOException;
import java.io.FileReader;
import java.io.FileWriter;

import java.util.*;
import java.util.stream.LongStream;

public class Main {
    // 標準入力
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    // 標準入力数値配列用 int
    static int[] inputval() throws Exception {
        String[] strarray = br.readLine().trim().split(" ");
        int[] intarray = new int[strarray.length];
        for (int i = 0; i < intarray.length; i++) {
            intarray[i] = Integer.parseInt(strarray[i]);
        }
        return intarray;
    }

    /* 標準入力数値配列用 long */
    static long[] inputLongArr() throws Exception {
        String[] strarray = br.readLine().trim().split(" ");
        long[] longarray = new long[strarray.length];
        for (int i = 0; i < longarray.length; i++) {
            longarray[i] = Long.parseLong(strarray[i]);
        }
        return longarray;
    }

    // 標準入力数値リスト用 int
    static List<Integer> inputIntList() throws Exception {
        List<String> strList = Arrays.asList(br.readLine().trim().split(" "));
        List<Integer> intList = new ArrayList<Integer>();
        for (String elem : strList){
            intList.add(Integer.parseInt(elem));
        }
        return intList;
    }

    // 標準入力数値配列用 integer 降順ソート用
    static Integer[] inputvalInteger() throws Exception {
        String[] strarray = br.readLine().trim().split(" ");
        Integer[] intarray = new Integer[strarray.length];
        for (int i = 0; i < intarray.length; i++) {
            intarray[i] = Integer.parseInt(strarray[i]);
        }
        return intarray;
    }

    /*標準入力long*/
    static long inputLong() throws Exception {
        return Long.parseLong(br.readLine());
    }

    /*標準入力long*/
    static int inputInt() throws Exception {
        return Integer.parseInt(br.readLine());
    }




    public static void main(String[] args) throws Exception {
        // write your code here
        int n = inputInt();
        long [] al = inputLongArr();
        int kCnt = 0;
        for (long a : al){
            if (a % 2 == 1){
                kCnt++;
            }
        }
        if (kCnt %2 == 0){
            System.out.println("YES");
        }else{
            System.out.println("NO");
        }

    }

}