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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	Scanner sc = new Scanner(System.in);
	public static void main(String[] args){
		new Main();
	}
	
	public Main(){
		new Aoj1130().doIt();
	}
	class Aoj1130{
		void doIt(){
			int N = sc.nextInt();
			int[] A = new int[N];
			int[] B = new int[N];
			int cnt = 0;
			for(int i = 0; i < N; i++){
				A[i] = sc.nextInt();
				if(A[i] % 2 == 0){
					B[i] = 1;
				}else{
					B[i] = 0;
				}
			}
			
			while(true){
				for(int i = 0; i < N; i++){
					if(B[i] == 1){
						for(int k = i + 1; k < N; k++){
							if(B[k] == 1){
								A[i] = A[i] + A[k];
								A[k] = -1;
								if(A[i] % 2 == 0){
									B[i] = 1;
								}else{
									B[i] = 0;
								}
							}
						}
					}
					if(B[i] == 0){
						for(int k = i + 1; k < N; k++){
							if(B[k] == 0){
								A[i] = A[i] + A[k];
								A[k] = -1;
								if(A[i] % 2 == 0){
									B[i] = 1;
									B[k] = -1;
								}else{
									B[i] = 0;
									B[k] = -1;
								}
							}
						}
					}
					
					//for(int k = 0; k < N; k++){
					//	System.out.print(A[k] + " ");
					//}
					//System.out.println("");
				}
				
				cnt = 0;
				for(int k = 0; k < N; k++){
					if(A[k] != -1)cnt++;
				}
				if (cnt >= 2) break;
			}
			if(cnt >= (N-1))System.out.println("YES");
			else System.out.println("NO");
		}
	}
}