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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

long long num[100001];
long long f[100001];

int main(void){
	long long n, i, j, k, cnt = 0;
	
	scanf("%lld", &n);
	
	for (i = 0; i < n; i++){
		scanf("%lld", &num[i]);
	}
	
	for (k = 0; k < n-1; k++){
		j = 0;
		for (i = 0; i < n; i = j+1){
			if (f[i] == 0){
				for (j = i+1; j < n; j++){
					if (num[i]%2 == num[j]%2){
						f[j] = 1;
						num[i] += num[j];
					}
				}
			}
			else if (i > 0) j++;
		}
	}
	
	for (i = 0; i < n; i++){
		if (f[i] == 0) cnt++;
	}
	
	if (cnt == 1) printf("YES\n");
	else printf("NO\n");
	
	return 0;
}