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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <map>
#include <string>
#include <stdio.h>
#include <math.h>

using namespace std;

#define min(a,b) ((a) < (b)?(a):(b))
#define max(a,b) ((a) > (b)?(a):(b))

map<char, int>mp;

int main()
{	
	int n,odd = 0;
	cin>>n;
	int arr[n];
	for(int i=0;i<n;i++)
	{	
		cin>>arr[i];
		if(arr[i] % 2)
			odd++;
	}
	if(odd % 2)
		cout<<"NO";
	else
		cout<<"YES";
	return 0;	
}